Lecture Notes on Slice Knots

Home , Claimed, Hopf link, Knot complement, Ribbon knot, Slice knot, Torus knot

Slice Knots: Knot Theory in the 4th Dimension

Peter Teichner

January 20, 2011

These notes are based on handwritten notes made by Justin Roberts from a lecture course given by Peter Teichner in San Diego in 2001. They were typed up by Julia Collins and Mark Powell who annotated the original version.

1 The Next Best Thing to the Unknot

Deﬁnition 1.1. A knot is an oriented, locally ﬂat, embedding of S1 ,→ S3.

K In traditional knot theory, a knot S1 ,→ S3 is trivial if it bounds a disc. Now, any knot S1 ,→ S4 is trivial. This is because the freedom of the extra dimension allows one to pass strands through one another and thus untie any knot. However this operation requires that the disc which this 4 3 4 4 4 unknotting operation traces out lives in the D on either side of S (S = D ∪S3 D ) - it is necessary to push one strand into the 4th dimension in one direction and the other strand into the 4th dimension the other way. The interesting question to ask is therefore whether a knot bounds a disc in D4 on one side only of S3. First we recall some deﬁnitions. Reminder. An immersion is a diﬀerentiable map whose derivative is everywhere injective. An embedding is an immersion which is also a homeomorphism onto its image, where the image has the subspace topology.

K Definition 1.2. An embedding S1 ,→ S3 is locally flat if for each point x ∈ S1 there is a neighbourhood U ⊂ S1 of x and a neighbourhood V ⊂ S3 of K(x) such that the pair (V,K(U)) can be mapped homeomorphically onto (D˚3, D˚1). q Similarly D2 ,→ D4 is locally flat if for each point x ∈ D2 there is a neighbourhood U of x and a neighbourhood V of q(x) such that the pair (V, q(U)) can be mapped homeomorphically onto (D˚4, D˚2). Definition 1.3. A knot K is slice if it is the boundary of a locally flat disc D2 embedded into the 4-ball D4.

2 4 We may also think of K as the cross-section of a locally ﬂat 2-sphere S in R by a hyperplane 3 R .

1 Remark 1.4. Flatness is essential. Any knot K ⊂ S3 is the boundary of a disc D2 embedded in D4, which can be seen by taking the cone over the knot (see ﬁgure below).

The cone over the knot is homeomorphic to a disc D2, and if we embed it into 4-space there will be no singularities. However, there is something displeasing about what is happen- ing at the vertex of the cone, where the knot gets squashed to a point. The problem is that the embedding at this point is not locally ﬂat; there is not a neighbourhood around it which looks topologically like the standard embedding of a disc D2 into D4.

Slice knots are special kinds of knots where it is possible to find such a disc whilst avoiding these kinds of singularities. The original motivation for the study of slice knots, and the first definition, was made by Fox and Milnor in 1958; they were interested in smoothing PL singularities of surfaces in 4-space, which arise naturally when considering complex hypersurfaces. Slice knots are also intimately related with the failure of the Whitney trick in 4-dimensions. The Whitney trick is used to remove intersections of submanifolds which cancel algebraically: if there are paths between the intersection points in each submanifold which form a loop, and this loop can be made to bound an embedded disk, then by isotoping across the disk the intersections can be removed. This works in high dimensions, but in dimension 4 the disks can only be immersed generically. The question of improving these to embeddings is like trying to slice a knot. Finally, slice knots are interesting because they enable us to make the set of all knots into a group, which will be our central object of study:

2 3 Example 1.5. If K is a knot which is symmetric with respect to a plane R ⊂ R then K is 4 2 slice because we can spin it through R+ about the axis R to produce the desired locally ﬂat 3 2 disc. [We can spin a point x = (x1, x2, x3, 0) of R+ about R according to the formula xθ = ∗ 4 (x1, x2, x3 cos θ, x3 sin θ). The spin K = {xθ : x ∈ K, 0 ≤ θ ≤ 2π} is a 2-sphere in R .] So if K is a knot in S3 and r : S3 → S3 is an orientation-reversing homeomorphism, and if K is the same knot but with the string orientation reversed, then K#rK is a slice knot.

Example 1.6. Our main bank of examples in this text will be the Twist Knots as shown below. The box with an n in it should really mean n − 1 full twists, so that the total number of twists is n. (Figure 1): We will see in due course that Kn is slice if and only if n = 0 or n = 2 (n = 0 is the unknot so is trivially slice).

How to see that a knot is slice We can visualise a slice disc by making movies. If a knot is (smoothly) slice then it bounds a disc D2 ⊂ D4 so that concentric 3-spheres move through (intersect) it to produce either

2 n

Figure-8: K General: K Trefoil: K-1 1 n

Figure 1: The Twist Knots

1. An ordinary nonsingular knot or link

2. A knot or link with singularities corresponding to one of

(a) Simple maximum or minimum (b) Saddle point

Maximum Minimum Saddle Point

Example 1.7. Stevedore’s knot, otherwise known as 61 in the standard knot tables, is the simplest slice knot (other than the unknot). The following “movie” shows how 3-spheres move through the slice disc:

The slice disc is shown schematically below - of course in reality this is a knotted disc in 4-space:

3 saddle point

minima

Example 1.8. Another example of a slice knot is the 8-crossing knot 88. Here is the corresponding slice ‘movie’:

Deﬁnition 1.9. A ribbon disc is a slice disc without local maxima. To construct a ribbon knot, start with an unlink (which corresponds to local minima; we can arrange the disk so that these come ﬁrst, then the saddles). Now add bands connecting them until the result is a knot/disc. Then local ribbon singularities are the only self-intersections, as opposed to clasp singularities. 2 2 3 The ribbon disc is the image α(D ) of a mapping α: D → R whose only singularities are of the following form. Each component of the singular set is the image of a pair of closed intervals in D2, one with endpoints on the boundary of D2 and one entirely interior to D2.

We can locally resolve a ribbon singularity into 4-space to get back a slice disc. See Figure 2, where the cross-hatched parts can be pushed oﬀ S3 into D4 to remove the self-intersections. This does not work for clasp singularities. Fact. Every ribbon knot is a slice knot. The following is a famous unsolved conjecture of Fox. Conjecture 1.10. Every slice knot is a ribbon knot. Note that this is about smooth knots. Any ribbon knot is smoothly slice, since we used Morse theory to get a handle decomposition of the slice disc. As we shall see below there are knots which are topologically slice but not smoothly slice, so in particular these cannot be ribbon.

4 Figure 2: Stevedore’s knot 61 bounds a singular disc with two arcs of intersection

2 The Knot Concordance Group

Deﬁnition 2.1. Knots K0 and K1 are called concordant if K0# − K1 is slice. (Here −K denotes the mirror image of the knot with reversed orientation, while # is connected sum.)

Deﬁnition 2.2. C := oriented knots in S3, # / ∼ where K0 ∼ K1 if they are concordant.

Notice that the group can be constructed by semigroup quotient, i.e. K1 ∼ K2 in C iﬀ there exist slice knots S1, S2 such that K1#S1 = K2#S2.

There is an easier to visualise picture of concordance which is related to cobordism of manifolds.

Deﬁnition 2.3. Two knots K0 and K1 are called concordant if there is a locally ﬂat embedding 1 3 3 3 of S × [0, 1] into S × [0, 1] having boundary the knots K0 and −K1 in S × {0} and S × {1} respectively.

So we can think of concordance as when two knots are connected by a cylinder embedded (nicely) into the 4th dimension. Slice knots, of course, are those which are concordant to the unknot.

Theorem 2.4. C really is a group.

Proof. For a knot K the knot rK is an inverse. It is possible to glue two null-concordances together, so that K1]K2 is slice if both K1 and K2 are slice. It is also the case that if K1 + K2 is slice and K2 is slice, then K1 is also slice. To see this glue the null-concordances together according to: K1 ∼ K1 + U ∼ K1 + K2 ∼ U. Lemma 2.5. K is slice if and only if there exists a ribbon knot R such that K#R is ribbon.

5 K

Unknot

Figure 3: Finding a ribbon knot within a slice cobordism.

Proof. “⇐” Ribbon knots are slice, so R is slice and K#R is concordant to K. Since K#R is ribbon and therefore slice, this means that K is concordant to a slice knot, and is therefore slice.

“⇒” Suppose K is slice. Then we can draw a cobordism to the unknot with the maxima ﬁrst, then saddles then minima In the middle of this cobordism we can ﬁnd a knot R which will be ribbon because it is cobordant to the unknot with only minima (Figure 3). Take the connected sum of R and K along the left-hand boundary of the concordance (Figure 4, left-hand picture). With a little imagination this can turn into the right-hand picture in Figure 4, which is a schematic of a ribbon disc.

Proposition 2.6. The group {knots, #} / {ribbon knots} is isomorphic to C.

Proof. Suppose K1 ∼ K2 in Knots/Ribbon. So there exist R1 and R2 ribbon such that K1#R1 = K2#R2. But ribbon knots are slice, so this means K1 ∼ K2 in Knots/Slice. Now suppose K1#S1 = K2#S2 for Si slice. By Lemma 2.5 we can ﬁnd ribbon knots Ri such that Si#Ri is ribbon for i = 1, 2. Then

K1#S1#(R1#R2) = K2#S2#(R1#R2) which we can re-bracket as

K1#((S1#R1)#R2) = K2#((S2#R2)#R1).

Since the addition of two ribbon knots is ribbon, we have the result that K1#ribbon = K2#ribbon. Thus K1 ∼ K2 in Knots/Ribbon.

6 K K

# #

R R

Figure 4: Seeing a ribbon knot in the connected sum.

We want to show that C is non-trivial. We can do this with additive knot invariants that vanish on ribbon knots (often more convenient to show).

3 Recall, the linking number of curves l1, l2 ⊂ S is deﬁned as lk(l1, l2) = l1 · F2, the intersection number of l1 with a Seifert surface F2 for l2. To generalise to 4 dimensions, choose Seifert surfaces 4 F1, F2 embedded in D transverse to the boundary such that ∂F1 = l1, ∂F2 = l2. Orient F1, F2 so that they induce the given orientations on l1 and l2. Then

lk(l1, l2) = F1 · F2 where this intersection number is taken in D4. The observation that linking information in S3 strongly corresponds to intersection data in D4 is a fundamental one, as we shall see.

We must show that this definition is independent of the Seifert surfaces we chose. Pick alternative 4 4 surfaces G1 and G2, in a different copy of D . Then glue the two copies of D together to get closed 2 4 ∼ surfaces F1 ∪ G1 and F2 ∪ G2 in S . Then H2(S ) = 0 so any intersections cancel: let R be the 3-chain whose boundary is F1 ∪ G1; then the intersections of F2 ∪ G2 with this are arcs whose endpoints are the intersections of F1 ∪ G1 and F2 ∪ G2. These endpoints have opposite intersection signs, so (F1 ∪ G1) ∩ (F2 ∪ G2) = 0. From this definition, we see that a link with non-zero linking number cannot be slice (i.e. bound disjoint discs in 4-space).

∼ 2g Let F be a Seifert surface for K. Then we have the Seifert pairing on H1(F ) = Z

S : H1(F ) × H1(F ) −→ Z;

7 S(x, y) = lk(x, y+), where x, y are representative chains of the homology classes, and y+ is the push oﬀ of y into S3 \ F along a normal vector to F . From S we can get the Alexander polynomial

T n A(K)(t) = det(S − tS ) ∈ Z[t]/ ± t .

We can show this is independent of basis choices and the choice of surface: this is the case since any two Seifert surfaces are related by some sequence of isotopies and handle additions/subtractions. By examining the eﬀect of such moves on the Seifert matrix, detailed below, one can see that the various invariants are independent of the choice of surface, without which property it would be hard to justify calling them invariant.

T Deﬁnition 2.7. The signature σ(K) is the signature of the symmetric form S + S , taken over R so that it can be diagonalised. This is also independent of choice of S.

Remark 2.8. Independence of the choice of S is from Morse theory – the elementary changes are handle additions which have the eﬀect  .  0 .    . .   S . .    S ←→  .   0 .     0 ··· 0 0 0  ········· 1 ∗

This is called S-equivalence. One also checks that the signature and the Alexander polynomial are invariant under integral congruences corresponding to changing the basis of H1(F ).

±1 Remark 2.9. We can normalise the Alexander polynomial A(K)(t) ∈ Z[t ] by 1. A(K)(t) = A(K)(t−1)

2. A(K)(1) = +1 (notice that S − ST is the intersection form J and has det ±1).

3. Addition of knots corresponds to block addition of Seifert matrices, so that the signature is additive and addition of knots corresponds to multiplication of Alexander polynomials: . A(K1]K2) = A(K1)A(K2).

1/2 −1/2 The Conway polynomial CK (z) is the z = t + t version of this.

Theorem 2.10. If K is slice then there exists a half-rank direct summand L in H1(F ) such that S|L = 0. Corollary 2.11. (a) K slice ⇒ σ(K) = 0.

8 (b) K slice ⇒ A(K) is of the form f(t)f(t−1) (up to t±n). Proof. (of corollary) 0 L (a) Let M = S + ST = , where N = C + CT . The matrix S − ST is non-singular LT N T T so S + S is non-singular over Z2, and therefore also over Z, that is det(S + S ) 6= 0, L−1 0 which means that det(L) 6= 0 and L is invertible over . Let P = . Then Q CL−1 −I 0 −I PMP T = , so σ(K) = σ(M) = σ(PMP T ) = 0. −I 0

0 A 0 A − tBT (b) S = so det(S−tST ) = det = det(A−tBT ) det(B−tAT ) = BC B − tAT C − tCT f(t)f(t−1) up to units.

Remark 2.12. The knot signature is not to be confused with the signature of a 4-manifold, which is the signature of the middle-dimensional intersection form on the 2nd homology of (some cover of) the manifold. (Although as we shall see for a judicious choice of 4-manifold the two signatures can coincide.) −1 1 −2 1 Example 2.13. RH Trefoil: S = , so M = S + ST = . Eigenvalues are 0 −1 1 −2 both negative, so σ = −2. 1 0 2 1 √ Figure-8: S = so M = . Eigenvalues are ± 5 so σ = 0. 1 −1 1 −2 The fact that σ = 0 for the Figure-8 agrees with the fact that σ is additive and the Figure-8 is amphichiral (so 2σ(K) = σ(K#K) = 0, since K#K is slice). 1 − t −t However, A(Fig-8) = det = −(1 − t)2 + t = −t2 + 3t − 1 =∼ −t + 3 − t−1, and 1 −1 + t this isn’t f(t)f(t−1) (since A(−1) is not a square). Therefore the Figure-8 is a 2-torsion element in C, and the Trefoil generates a free summand of C. Example 2.14. We return to the twist knots. We can get a Seifert surface by resolving the obvious disc with clasp singularity into two half twisted bands. Then, for the twist knot Kn we get a Seifert matrix −1 1 S = . 0 n

We constructed a homomorphism σ : C → Z, the signature, by using the theorem that any slice knot K with a Seifert surface F has a Lagrangian L ⊂ H1(F ) (i.e. a half-rank direct summand for which S|L = 0). Now we need to prove this.

9 n

Figure-8: K General: K Trefoil: K-1 1 n

Figure 5: The Twist Knots again.

Figure 6: Add a tube and feed the band through. The knot remains the same!

Proof of Theorem 2.10. Warmup exercise for ribbon knots: Prove for ribbon knots that σ is well-deﬁned. Actually, it is enough to show that σ vanishes on ribbon knots for some Seifert surface, because we know C = knots/slice =∼ knots/ribbon and σ is independent of choice of surface. Ribbon singularities can be resolved (see Figure 6). Desingularising the whole ribbon like this gives a disc and a tube for each ribbon singularity, and with obvious Lagrangian: circles surrounding the ribbon cuts; a basis for the Lagrangian is given by an unlink of circles.

Proof of Theorem: Step 1: There exists an oriented submanifold M 3 ⊂ D4 with boundary F ∪∆, where F is a Seifert surface of K and ∆ is a slice disc.

Proof of Step 1. This elementary application of obstruction theory was lifted from the text- book of Lickorish. Let X be the exterior of K. We want to deﬁne a map φ: X → S1 so that 1 −1 φ∗ : H1(X) → H1(S ) is an isomorphism and φ (pt) = F . On a product neighbourhood of F in X, deﬁne φ to be the projection F × [−1, 1] → [−1, 1] followed by the map t 7→ eiπt ∈ S1. Let φ map the remainder of X to −1 ∈ S1.

Let N = ∆ × I2, a neighbourhood of ∆. We extend φ to the rest of ∂(D4 − N) so that the inverse image of 1 ∈ S1 is F ∪ (∆ × ∗) for some point ∗ ∈ ∂I2 (note: ∂D × ∗ is a longitude of K). We now need to extend the map over all of D4 − N.

Consider the simplices of some triangulation of D4 − N. Let T be a tree in the 1-skeleton

10 containing all the vertices of this triangulation, that contains a similar maximal tree of ∂(D4 − N). Extend φ over all of T in an arbitrary way. Then on a 1-simplex σ not in T , deﬁne φ so that if c 4 is a 1-cycle of H1(D − N) consisting of σ summed with a 1-chain in T (joining up the ends of σ), 1 then [φc] ∈ H1(S ) is the image of [c] under the isomorphism

∼ 4 = φ∗ 1 H1(D − N) ←− H1(X) −→ H1(S ).

4 4 Trivially, the boundary of a 2-simplex τ of D − N represents zero in H1(D − N), so [φ(∂τ)] = 1 0 ∈ H1(S ). Hence φ is null-homotopic on ∂τ and so extends over τ. Finally, φ extends over the 3 & 4-simplices, as any map from the boundary of an n-simplex to S1 is null-homotopic when n ≥ 3.

Now regard φ: D4 − N → S1 as a simplicial map to some triangulation of S1 in which 1 is not a vertex. Then φ−1(1) is a 3-manifold M 3, and φ was constructed so that ∂M 3 = F ∪ (∆ × ∗).

Step 2: P := ker[H1(∂M; Q) → H1(M, Q)] is a Lagrangian subspace of dimension g (where ∂M has genus g).

Proof of Step 2. A Lagrangian subspace is a vector subspace P of half rank which satisﬁes P = P ⊥. Look at the homology exact sequence of (M, ∂M) (over Q):

0 / H3(M, ∂M) / H2(∂M) / H2(M) / H2(M, ∂M) /

P / 0 o7 ooo k ooo ooo ooo H1(∂M) / H1(M) / H1(M, ∂M) / H0(∂M) / H0(M) / 0

∼ k ∼ k ∼ Lefschetz duality says H1(M, ∂M) = H2(M) (since H (M, ∂M) = Hn−k(M) and H (M, ∂M) = Hk(M, ∂M) by the Universal Coefficient Theorem, as we are working over a field). Poincaréduality k ∼ says H (M) = Hn−k(M), which again implies dim(Hk(M)) = dim(Hn−k(M)). So let

a = dim(H3(M, ∂M)) = dim(H0(M))

b = dim(H2(∂M)) = dim(H0(∂M))

c = dim(H2(M)) = dim(H1(M, ∂M))

d = dim(H2(M, ∂M)) = dim(H1(M))

e = dim(H1(∂M))

11 Then the exactness of the sequence implies that

a − b + c − d + e − d + c − b + a = 0 ⇒ 2(a − b + c − d) + e = 0

But dim P = e − d + c − b + a = e + (a − b + c − d). 1 Therefore 2(dim P − e) + e = 0 ⇒ 2 dim P = e, and thus dim P = 2 dim(H1(∂M)). ∼ Now note that H1(∂M) = H1(F ) (recall ∂M = F ∪ ∆ and H1(∆) = 0). Suppose we have α, β ∈ ker[H1(∂M) → H1(M)]. There exist surfaces A, B ⊂ M with ∂A = α, ∂B = β. When α is moved to α+, the surface A can also be moved oﬀ M to M × {1} (so ∂A+ = α+), and then the intersection of A+ and B is empty. Thus lk(α+, β) = A+ · B = 0.

To move from Q to Z coeﬃcients is no problem, although the Z-kernel of H1(F ) → H1(M) might not be a direct summand. Use instead L = {a ∈ H1(F ): ∃ n ∈ Z\{0} s.t. na ∈ P }. Its rank over Q is the same as that of the kernel and it is a direct summand because H1(F )/L is torsion-free - any torsion is automatically in L. Note also that S(na, b) = 0 =⇒ S(a, b) = 0 by linearity, so S|L = 0, as required.

Now we can use the twist knots Kn to show that C is not ﬁnitely-generated.

1 Deﬁnition 2.15. For ω ∈ S \{1} ⊆ C, we deﬁne the twisted signatures σω : C → Z by

T σω(K) := σ((1 − ω)S + (1 − ω)S ) where S is a Seifert matrix for K.

Notice that the matrix Q := (1 − ω)S + (1 − ω)ST is hermitian so the eigenvalues are real and T 2g the signature is well-defined. Note also that Q = (1 − ω)(S − ωS ) and det Q = (1 − ω) AK (ω) where AK is the Alexander polynomial. The finitely many zeros of the Alexander polynomial mean finitely many places at which the form becomes degenerate. In fact, σω(K) is continuous as a function of ω (K fixed) except at zeros of the Alexander polynomial. Since signature is an integer, this means that the signature is constant between roots and jumps around the unit circle.

Theorem 2.16. The signature σω : C → Z is a homomorphism (i.e. σω vanishes on slice knots and is additive).

0 ∗ Proof. K slice implies that Q = . The proof is thus the same as before provided that ∗ ∗ AK (ω) 6= 0. If AK (ω) = 0 then we deﬁne σω to be the average of the two limits, extending the deﬁnition to the bad points.

The following Theorem is the central concrete aim of the rest of these notes.

Theorem 2.17. (a) Kn slice ⇔ n = 0, 2 (This was ﬁrst proved by Casson-Gordon)

12 2 (b) {Kn} are independent in C for n < 0 and for (4n + 1) = l (n > 0)

Corollary 2.18. All Kn are distinct in C except for the unknot K0 = K2.

Conjecture 2.19. All Kn are Z-independent (other than K0 = K2 = 0) in C. Proof of Theorem 2.17. Developing the tools to prove the whole of this Theorem will be the ulti- mate aim of these notes. For now we give the proofs for which we currently have the technology. −1 1 We have σ : C → deﬁned by σ((1−ω)S +(1−ω)ST ), where for twist knots K , S = . ω Z n 0 n

Compute σω(Kn): C → Z; this will jump at the roots of the Alexander polynomial. −1 + t 1 A (t) = det = n(−1 + t)(1 − t) + t = −nt2 + (2n + 1)t − n Kn −t n − nt 2 2n+1 = −n(t − n t + 1) This has roots q 4n2+4n+1 √ 2n + 1 2 − 4 2n + 1 ± 4n + 1 ± n = . n 2 2n

So AKn has real roots if and only if n > 0, in which case σ is constant (σ ≡ 0) around the unit circle. For n < 0, the roots are on the unit circle, since ωn · ωn = 1, and they are all distinct from one another. −2 1 Computing at t = −1 we get σ = −2. 1 2n (N.B. If A is a symmetric n × n matrix and the principal minors A1,A2,...,An are all non-zero then the signature of A is n − 2×(the number of changes of sign in the sequence 1,A1, ..., An).) Therefore the {Kn} are independent in C for n < 0. (We have now proved half of (b)).

Now consider n > 0. The Alexander polynomial is reducible over Q if and only if 4n + 1 is a square. So assume it is not a square. ±1 For any symmetric irreducible polynomial p ∈ Q[t ], deﬁne a homomorphism hp : C → Z2 by ±1 K 7→ {exponent of p in Ak, mod 2} (factorised over P.I.D. Q[t ]). This is additive under # because Alexander polynomials multiply. We need to show that it −1 −1 vanishes on slice knots. But for these, Ak(t) = f(t)f(t ), so the exponent of p(t) = p(t ) is even −1 in it. For case (c) we have AKn = −nt + (2n + 1) − nt , which are distinct irreducible symmetric polynomials. By applying the appropriate hp function we can show that Kn are Z2-independent. This proves (c).

13 The hard case to do is “4n + 1 = square”, in (b). The conjecture that all {Kn}6=0,2 are Z- independent in C is partially known by work of Livingston and Naik.

Suppose 4n + 1 = l2. Then 4n = (l − 1)(l + 1), l = 2m + 1, so n = m(m + 1), m > 0. Notice that Kn has a genus 1 Seifert surface F , so H1(F ) has two basis vectors, say s and l.

For m = 1 we have K2, called by mathematicians the Stevedore’s knot, which is actually slice. The claim is proved by ﬁnding that γ = −s + l is a vector with square 0:

−1 1 −1 −1 1 = 0 0 2 1

The curve on the Seifert surface is the unknot - visibly it is a 0-framed (i.e. self-linking zero) unknot.

K2 is therefore ribbon: we see this by cutting out an annulus around our curve γ = −s + l to get two parallel copies of γ. This is possible since γ has zero self-linking. We then attach slice discs for γ and γ0, and this will introduce only ribbon singularities. To do the surgery and get two non-intersecting discs we need a 0-framing on the unknotted curve, and we also need the curve γ to be itself slice. Lemma 2.20. If n = m(m + 1) then there are precisely two curves with self-linking zero. These are γ = (−m)s + l and γ = (m + 1)s + l.

Proof. Let γ = αs + βl, and suppose lk(γ, γ+) = 0. Then we have:

0 = lk(γ, γ+) = α2 lk(s, s+) + αβ(lk(s, l+) + lk(l, s+)) + β2 lk(l, l+)

= −α2 + αβ + m(m + 1)β2 Completing the square with respect to α, we get that β ± β(2m + 1) α = 2 so that γ = (m + 1)s + l or γ = (−m)s + l.

14 Deﬁnition 2.21. K is called algebraically slice if there exists a Lagrangian in H1(F ). Remark 2.22. (a) K slice ⇒ K algebraically slice, by Theorem 2.10.

(b) The Km(m+1) are algebraically slice by Lemma 2.20.

For m = 2 (n = 6) we consider γ = −2s + l. This is in fact the trefoil, which is not slice; we can calculate its signature or show that its Alexander polynomial does not factorise as it should. Similarly for Km(m+1) we get γ = T (m, m + 1), a torus knot which twists m times round the meridian of a standardly embedded torus in S3, and m + 1 times around the longitude. Such torus knots are slice if and only if m = 1; again one can calculate their signatures. This gives us a good reason to believe that these twist knots are not slice, though we will need the following theorem before we can actually prove it.

Theorem 2.23. (Cochran-Orr-Teichner) If K is a genus 1 knot with Ak 6= 1 then K slice ⇒ there R exists γ ⊂ F such that S(γ, γ) = 0 and γ is a generator of H1(F ) such that ω∈S1 σω(γ) = 0. So by integrating the twisted signatures of the torus knots T (m, m + 1) around the circle we can obstruct the sliceness of the twist knots. We will prove this theorem in Section 6, once we have developed the necessary machinery. This is our motivating example, although of course the new theory has a far wider reach than merely reproving the results of Casson and Gordon. It is not known whether every slice knot has a (collection of) slice γ-curve(s), but this result is a substitute. The original proofs of the parts of this Theorem were done by Casson-Gordon for part (a), while for (b) Tristram did the case n < 0 and Jiang the case 4n + 1 prime. The remaining statements are contained in the work of Cochran-Orr-Teichner.

3 A Survey of Homology, Intersection Forms and Linking Forms in Low Dimensions

3.1 Homology

Deﬁnition 3.1. Let X be any space. Ωi(X) is the group of oriented bordism classes of manifolds i i of dimension i. Elements of Ωi(X) have the form (M , f), where M is a closed oriented manifold i and f : M → X is a continuous map. We say that two such pairs (M0, f0), (M, f) are bordant if the disjoint union M0 t −M is the boundary of an oriented (i + 1)-manifold W , and there exists a continuous mapping h: W → X such that h|M0 = f0, h|M = f. The group operation is disjoint union, and inverses are by reversing orientation.

15 Proposition 3.2. Let X be any space; i = 0, 1, 2, 3. Then Ωi(X) is isomorphic to Hi(X), with the isomorphism given by (M, f) 7→ f∗[M]

2 Proof. The ﬁrst ‘essential” singularity in an oriented manifold is a cone on CP . (With Z2, a cone 2 on RP ) n Proposition 3.3. Let X be a manifold of dimension n ≤ 4. Then homology classes Hi(X) are represented by (embedded) submanifolds

3.2 Intersection and Linking Pairings Let Xn be a closed oriented manifold. Then the Universal Coeﬃcient Theorem gives:

0 → Ext1 (H (X), ) → Hp(X) → Hom(H (X), ) → 0. Z p−1 Z p Z By Poincar´eDuality we have ∼ p Hn−p(X) = H (X). We also have the isomorphisms1

Ext1 (H (X), ) =∼ Hom(TH (X), / ) =∼ THp(X) Z p−1 Z p−1 Q Z where TA = {x ∈ A | sx = 0 for some s 6= 0 ∈ Z}. So rewrite as:

p 0 / Hom(THp−1(X), Q/Z) / H (X) / Hom(Hp(X)/T Hp(x), Z) / 0 O O O

0 / THn−p(X) / Hn−p(X) / Hn−p(X)/T Hn−p(X) / 0 Thus

(a) We get an intersection form IX : Hn−p(X)/T Hn−p(X) × Hp(X)/T Hp(X) → Z

(b) We get a linking pairing lkX : THn−p(X) × THp−1(X) → Q/Z which are both non-singular.

Geometrically these pairings are given by counting intersections (with sign). The linking number is got by taking multiples of one class, getting a bounded guy, intersecting and dividing: A ∩ β lk (A ,B ) = # n−p p ∈ / X n−p p−1 m Q Z 1See Appendix 6 for proof.

16 where ∂βp = mBp−1(m 6= 0). (This is independent of m.)

These are particularly interesting in ‘middle’ dimensions. i.e. (a) p = n − p, so n = 2p is even

(b) n − p = p − 1, so n = 2p − 1 is odd. Then symmetry is that

p (a) IX (Ap,Bp) = (−1) IX (B,A)

p−1 (b) lkX (ap, bp) = (−1) lkX (a, b). Thus, for n = 4 there is a symmetric intersection form; for n = 3 there is a symmetric linking form. However, for n = 4 there is also an interesting linking form for manifolds with boundary which we shall also make use of later.

For manifolds with boundary, which are compact and oriented, we look at the relative (co)homology groups in the upper short exact sequence above. Thus:

IX : Hn−p(X)/T Hn−p(X) × Hp(X, ∂X)/T Hp(X, ∂X) → Z

lkX : THn−p(X) × THp−1(X, ∂X) → Q/Z

We we cannot deal with symmetry here. But we can at least use Hp(X) → Hp(X, ∂X) to get a nicer pairing. Remark 3.4. Relative homology classes are representable by proper submanifolds of (M, ∂M). Example 3.5. Let W 4 be a compact oriented 4-manifold with boundary ∂W = M. Then long exact sequence of a pair in homology gives us:

() H2(W ) → H2(W, M) −−→ H1(M) → H1(W ) whilst Lefschetz duality and the Universal Coeﬃcient Theorem give us

∼ 2 (∗) H2(W, M) = H (W ) −−→ Hom(H2(W ), Z) → 0.

Assume for now that H1(W ) = 0; then (∗) is an isomorphism, as the relevant Ext group from the Universal Coeﬃcient Theorem vanishes, and () is a surjection. We have the exact sequence

IW ∗ H2(W ) −−→ (H2(W )) → H1(M) → 0.

4 Concrete example: W = B ∪framed link L 2-handles.

17 Think of 2-manifolds: 1 handles attached along framed links S0 in boundary S1.

B2 x B2

S1 x B2

2 2 3 4 Attach 2-handles B × B to framed links in S = ∂B . Use Z to give the framing of each component of the link (write a number next to it in a diagram) , i.e. to specify the thickening of each S1 ,→ S3 to an embedding S1 × B2 ,→ S3. The framing measures the self-linking of the boundaries of the core B2 × {0} of each 2-handle by measuring the linking of two circles 2 1 1 2 ∂B × {pi} = S × {pi} ⊂ S × B , i = 1, 2. ∼ n ∼ ∼ We get a linking matrix V . W ' 0-cell ∪ n 2-cells, so H2(W ) = Z , π1(W ) = 0 = H1(W ). We have:

n V =IW n Z −−−−→ Z → H1(M) → 0 2 A basis for H2(W ) can be represented by the cores B × {0} of the 2-handles, union with cones on link components, or Seifert surfaces for the components. Use cones, intersected with the Seifert surfaces, to compute linking numbers as intersections. Now, return to relating the forms algebraically. The linking form λ on M is determined by IW . Take a torsion element a in H1(M)(λ is defined on the torsion part of H1(M), so we may as well assume that IW has finite cokernel, so all of H1(M) is torsion). We can lift a, b to elements A, B ∈ H2(W, M), since () above is a surjection. Now find an integer m such that mB 7→ mb = 0 ∈ H1(M), and so mB comes from an element C of H2(W ). Then take the intersection of C and A using the image of A under (∗), and divide by m. We thus obtain λ(a, b) If we have: n IW n Z −−→ Z → cokernel = H1(M) → 0

18 M a b

we get a pairing: −1 (a, b) ∈ H1(M) 7→ (ea, IW (eb)) n where IW is inverted over the rationals, tilde denotes a lift into Z , and the brackets denote the Kronecker pairing i.e. dot product of vectors. Compare with the localisation exact sequence in L-theory. Let A be a Principal Ideal Domain. Then the group Ln(A) is the group of algebraic cobordism classes of chain complexes over the ring A. Ln(A, A − {0}) is the group of cobordism classes of such chain complexes which become contractible upon inverting all non-zero elements of A, which is the for n = 4 is the group of linking forms on the 1st homology of the chain complexes. L4(A) is the group of non-singular intersection forms on 4-dimensional chain complexes. The L-theory localisation exact sequence is:

4 4 ∂ 4 3 · · · → L (Z) → L (Q) −→ L (Z, Z − {0}) → L (Z) → · · ·

3 3 Every 3-manifold is null-cobordant (Theorem 3.6), so M ∈ L (Z, Z − {0}) maps to zero in L (Z). 4 It therefore lifts to L (Q); the linking form on M is given by intersections in a 4-manifold which it bounds, with a non-singular form after localisation, in other words allowing rational numbers to invert the matrix. This corresponds to the homology of M being Z-torsion. Recall: given a framed link L ⊆ S3, we attach handles and denote the resulting 4-manifold by 4 3 4 3 3 WL, and ML := ∂WL. ML comes from doing the surgery on S speciﬁed by the link, and W is a null-cobordism for M.

Theorem 3.6 (Bing, Rochlin, Lickorish, Wallace, Rourke). Any closed oriented 3-manifold is the boundary of a 4-manifold, i.e. Ω3 = 0. Proof. We use the Pontrjagin-Thom construction to get the result that

Ω ∼ π (MSO) := lim π (MSO(k)) 3 = 3 −→ 3+k

19 where MSO(k) is the Thom space of the universal bundle over the classifying space BSO(k): it is the Thom spectrum for the generalised homology theory of oriented bordism. Ω1 and Ω2 are zero by classiﬁcation of 1- and 2-manifolds, so therefore π1+k(MSO(k)) and π2+k(MSO(k)) are zero for k suﬃciently large. We also know that MSO(k) is a (k−1)-connected space since it is the Thom space ∼ of a CW-complex, so that πm(MSO(k)) = 0 for 1 ≤ m < k. Also πk(MSO(k)) = Hk(MSO(k)) ∼ by the Hurewicz Theorem, so that π3(MSO) = H3(MSO).

∼ Now by the Thom isomorphism theorem, Hi+k(MSO(k)) = Hi(BSO(k)). By deﬁnition of ∼ BSO(k) we have πn(BSO(k)) = πn−1(SO(k)).

∼ Therefore: π1(BSO(k)) = π0(SO(k)) = 0 since SO(k) is connected. Then π2(BSO(k)) = ∼ H2(BSO(k)) by Hurewicz. Next π3(BSO(k)) = π2(SO(k)) = 0 since SO(k) is a Lie group. Applying Hurewicz again tells us that the map π3(BSO(k)) → H3(BSO(k)) is a surjection, which means H3(BSO(k)) = 0. ∼ Remark 3.7. We can continue to get Ω4 = Z. Then, step 2 is to surger the 4-manifold on circles to kill π1. (The circles are embedded, with neighbourhoods which are trivial bundles.) This is equivalent to exchanging the dots for zeros in a Kirby diagram; turning the handle decomposition upside-down and repeating with the 3-handles we get only 0-handles and 2-handles. The cobordism class of a 4-manifold is given by the signature of the intersection form on H2(W ; Z) - for a proof of this see for example Kirby’s Topology of 4-manifolds spin Addendum: we can assume all framings are even. This comes from Ω3 = 0, some handle- +1 2 slides, and the Wu formula. e.g. L = ; WL = CP − ball (which is the Hopf disk bundle over S2. See Gompf and Stipsicz p106), is not spin, but it can be closed oﬀ with a 4-ball, so it isn’t a counterexample. The Wu formula says that the second Stiefel-Whitney class w2(W ) is characterised by w2 ∪ x = x∪x mod 2. Since every 3-manifold is parallelisable, every 3-manifold can be given a spin structure, and every spin 3-manifold is the boundary of a spin 4-manifold, such a spin 4-manifold must have 2 x ∪ x = 0 mod 2 for all x ∈ H (W ; Z). There must therefore be, by Kirby’s theorem, a sequence of handle slides which leaves each component of the link having even framing. (See Section 4.2 below.)

Corollary 3.8 (Kervaire, in higher dimensions). Any knot S2 ⊆ S4 is slice; i.e. extends over a 3-ball in B5.

Proof. Pick a Seifert surface F 3 ⊆ S4, ∂F 3 = K. (Use the standard construction of a Seifert surface using a map S4 − K → S1 representing a generator of H1(S4 − K): Alexander Duality 1 4 2 ∼ 3 ⇒ H (S − S ) = Z). By Theorem 3.6, F is obtained from surgery on B (cap it oﬀ with another B3 to get S3, use the theorem, and remove at the end.) Therefore the converse is also true; we can surger F to become B3 along some link L ⊆ F . Now it’s no problem: we need only to surger along these by adding 2-disks in B5. Each circle of L bounds a disk in B5, and they are disjoint, purely by a general position dimension count.

20 We now need to discuss framings: thicken each B2 ⊆ B5 up to a B2 × B2. We need even framing on the link to do it right. Recall that the normal bundle of a disc B2 in B5 is trivial, but then can 3 2 always ﬁnd in any trivial R -bundle on S a 2-plane sub-bundle with any even Euler characteristic. Thus, surgeries are possible ambiently, if and only if we have even framings, but this is always possible so we can surger F 3 ambiently in B5 to become B3.

3.3 High dimensional Concordance Recall: {S1 ⊆ S3} C = . 1 ∂{B2 ⊆ B4} There is an analogous deﬁnition of high dimensional concordance groups:

{Sn ⊆ Sn+2} C := n ∂{Bn+1 ⊆ Bn+3}

Remark 3.9. Codimension 2 is the only interesting case to define such concordance groups, at least in Piecewise-Linear theory: we have the Schönfliestheorem for codimension 1. The difficulty for C1 corresponds to π1 being non-trivial. The proof of unknotted-ness in codimension 3 starts by observing that the complement has the same π1 as an unknot complement.

Theorem 3.10. For n ≥ 4, the groups Cn are 4-periodic in n, in the PL category;

(a) C0 = C2 = 0;

(b) C1 is unknown;

(c) C2k = 0 k ≥ 2; ∼ − (d) C4k+1 = AC k ≥ 1 ∼ + (e) C4k−1 = AC k ≥ 2 In addition: + 0 → C3 → AC → Z2 → 0 σ(S+ST ) with the Z2 term coming from Rochlin’s theorem, as 8 (mod 16). Recall the deﬁnition of the algebraic concordance groups:

{Seifert forms S : n → ( n)∗ with S ± ST an isomorphism} AC± = Z Z {metabolic forms: those with a Lagrangian summand}

21 Denotes the unknot.

Remark 3.11. AC− was original group, (k = 0); AC is the appropriate thing in the other dimensions. ± The map: Cn → AC given by the Seifert form is clear. ± L L L AC are in fact isomorphic to ∞ Z ⊕ ∞ Z2 ⊕ ∞ Z4. − Lemma 3.12. C1 → AC is onto. Proof. Create a knot by making a banded Seifert surface for it: just twist the bands and link them appropriately. Given any matrix S such that S − ST = J, where J is the standard symplectic matrix with block diagonals of the form

0 1 , −1 0 we can do it. As J is the only skew isomorphism (of a given rank, up to congruence), we are done. (In higher dimensions, use plumbing: plumbing of disks creates linking of their boundary spheres.)

Sketch proof of theorem 3.10 on Cn≥4. We have already proved C2 = 0, while C0 = 0 is clear. Pick a Seifert surface for the knot. By ambient surgeries below the middle dimension, we can assume that F is alterable to a slice disk Bn+1 ⊆ Bn+3 for n even. This finishes case (c). If instead S2k−1 ⊆ S2k+1, with S2k−1 ⊆ F ⊆ S2k+1, then we may assume that F is a connected sum: ](Sk × Sk) − B˙ 2k by again performing surgery below the middle dimension. Note that the first stage of this is to kill π1(F ); so that the fundamental group of the knot complement is infinite cyclic; as intimated above this is one of the places the argument goes catastrophically wrong in the classical dimension. After this concordance we have an analogue of the ](S1 × S1) − B˙ 2 which occurs in the dimension 3 case. Now S is again the linking form on the middle homology of this subset F inside S2k+1. ± Arrange as before to get the onto case AC as appropriate. Final step: if S has a Lagrangian in 2k Hk(F ), then the original knot was slice: this is true when k > 1. In order to see this, represent the Lagrangian by a link L = q Sk ⊆ F 2k ⊆ S2k+1. For k ≥ 3 use the Whitney trick to make them disjoint. For k = 2 we need the stable Whitney trick, and the power of transverse spheres. We need a lemma to complete the proof:

22 Lemma 3.13. If L: q Sk → S2k+1 has trivial linking numbers and k > 1, then L is slice (in fact trivial). We are then done as normal by surgery. The classical dimension is diﬀerent because the dimension does not drop with this progression: e.g. when trying to slice a knot S3 ⊆ S5, at this stage have to slice S2 ⊆ S5.

Proof of Lemma: We have to remove intersections of discs Bk+1 ⊆ B2k+2. Since the linking numbers of the boundaries vanish, we know that any intersections pair up, so we can use the Whitney trick to cancel them. The ambient dimension is at least 6, so this works admirably.

Remark 3.14. Addendum to high dimensional concordance theorem. It is very important to specify the category in we which we work, either: Smooth, Piecewise Linear, or Topologically Locally Flat. Theorem 3.10 applied to PL version. For n ≥ 4,  0 n even PL TOP  + Cn = Cn = AC n = 4k − 1 AC− n = 4k + 1, whereas in the smooth case there is a problem with realising symmetric forms using plumbing: can DIFF get exotic spheres knotted. So we have the same result for Cn for n 6= 4k ± 1, but:

DIFF + 0 → C4k−1 → AC → Znk → 0

+ The map AC → Znk is given by: σ(S + ST ) S 7→ . 8 nk is the order of the Kervaire-Milnor exotic sphere Σ4k−1 in the group of exotic spheres Θ4k−1. Recall that: {manifolds homeomorphic to S4k−1} Θ := , 4k−1 {those diffeomorphic to S4k−1} with the group operation of connected sum is a finite group. In order to prove that Σ]Σ is diffeomorphic to S4k−1, use the h-cobordism Σ × I minus a tube (the connected sum) and a ball (so we have a cobordism to the standard sphere).

Remark 3.15. In the 4k + 1 case there is a similar Arf invariant problem, better known in this case as the Kervaire invariant problem.2.

DIFF − 0 → C4k+1 → AC → {Z2 or 0} → 0

2It is now known, by recent work of Hopkins, Hill and Ravenal, that no manifold exists with Kervaire invariant 1 in dimensions greater than 126, while in some dimensions below there is such a manifold; 126 is still open.

23 The problem in the previous proof was trying to realise an even unimodular form stabilised by hyperbolics geometrically.

Theorem 3.16 (Serre, 1962).

k T ∼ M (S + S ) ⊕ hyperbolic = E8 ⊕ hyperbolic i=1 with 8k = σ(S + ST ).

To realise this as an intersection form of a Seifert surface W 4k ⊆ S4k+1 do plumbing. By Serre’s theorem, it is suﬃcient to be able to realise the E8 form. This is equivalent to taking a link L of 4k−1 ∼ (2k − 1)-spheres in S (framed, with a π2k−1(SO(2k)) = Z worth of framings to choose from) whose linking matrix corresponds to the E8 form’s intersection matrix, and using this to attach handles to B4k as usual to make

4k _ 2k WL ∪ 2-handles ' S .

4k ∼ ]L Note that π1(WL) = 1, and the middle dimensional homology is H2k(WL ) = Z , with the intersection form: ∗ ∼ H2k(WL) → H2k(WL) = H2k(WL,ML) given by the linking matrix. The isomorphism here is a combination of Universal coeﬃcient theorem and Poincar´e-Lefschetz duality. The boundary ML of WL also has trivial fundamental group, and is a homology sphere if and only if the linking form is non-singular. This can be seen from the long exact sequence of the pair (WL,ML). The non-trivial part of this sequence is

H2k+1(WL,ML) → H2k(ML) → H2k(WL) → H2k(WL,ML) → H2k−1(ML) → H2k−1(WL) which yields ]L φ ]L ∗ 0 → H2k(ML) → Z −→ (Z ) → H2k−1(ML) → 0. If the linking form is non-singular then the central arrow φ is an isomorphism which implies the vanishing of the middle-dimensional homology groups of M shown. For k > 1, the h-cobordism theorem, which implies the high-dimensional Poincar´ehypothesis, then implies that we have a 4k−1 homeomorphism ∂WL ≈ S . Note that WL is not an h-cobordism which would show this; one is however certain to exist. ∂WL is Σ4k−1, the Kervaire-Milnor sphere ∂(E8). (This is independent of the choice of link: it is classiﬁed up to isotopy by the linking form.) Also in this construction W 4k always embeds in S4k+1 (proof below), and is a Seifert surface and knot pair. This shows that

CAT + C4k−1 AC

4k−1 for CAT = PL or TOP, but not in the smooth case, because Σ4k−1 is not diﬀeomorphic to S .

24 Remark 3.17. n2 = 28; Σ7 generates Θ7. This is the famous initial example of Milnor. 4k+1 Proposition 3.18. WL ⊆ S . Corollary 3.19. The proposition, with k = 1, implies that any closed oriented 3-manifold embeds in S5. (Note that Whitney’s best case would give an embedding in S6 here.)

Proof of Proposition 3.18. Consider WL × I; each handle has its thickening increased by a dimension, so this is of the form B4k+1 ∪ (2k − handles). They now attach along a trivial link because of the dimension shift. Then, taking B4k+1 as one half of S4k+1, we can attach the 2k-handles ambiently in the other half, using the fact that the link is slice. We then have to make sure the framing is right to be sure that the slice disks can be thickened to 4k-dimensional 2k-handles B2k × B2k. However, at present we have trivially framed 4k +1-dimensional 2k-handles B2k ×B2k+1. We claim that within any 4k + 1-plane bundle, it is possible to ﬁnd 4k-bundles with any even Euler class. To see this, examine the exact sequence of homotopy groups associated to the ﬁbration SO(2k) → SO(2k + 1) → S2k:

2k π2k(S ) → π2k−1(SO(2k)) → π2k−1(SO(2k + 1)) → 0. ∼ 2k−1 4k−1 ∼ The framing of the link L = qS ⊆ S yields elements of π2k−1(SO(2k)) = Z as given by the Euler class of the vector bundle, while after allowing an extra dimension the, now trivial, link component’s framing gives elements of π2k−1(SO(2k + 1)). From the exact sequence, every element of π2k−1(SO(2k + 1)) lifts to π2k−1(SO(2k)). It is trivial precisely when the element of 2k π2k−1(SO(2k)) comes from an element of π2k(S ). Since this last group is generated by the identity 2k map, and maps to TS ∈ π2k−1(SO(2k)), which has Euler number 2, we see that trivial framings lift to even framings when we reduce the ambient dimension by one. It is precisely even framings on spheres which can be extended across disks; furthermore as we discussed above any 3-manifold can be represented by a link which has only even framings. This completes the proof. ∼ Theorem 3.20. If a 3-manifold M = ML, for L a 0-framed link which is the union of two slice links, then M 3 embeds in S4. Example 3.21. Two unknots linked n times via n twists, 0-framed. In other words, two ﬁbres of ∼ 2 n ∈ Z = π3(S ). 0 n H (M ) =∼ coker =∼ ⊕ 1 L n 0 Zn Zn 4 Actually ML is L(n, 1)]L(n, 1), but a single lens space does not embed in S . 4 Proof of Theorem 3.20. If the whole link were slice, it would be easy: WL ⊆ S as before, but with the ambient dimension one fewer. We need the zero framing here to be able to thicken the slice discs to 2-handles inside B4. Now if the individual components are slice, attach one 2-handle in 4 4 4 4 one B half of S , and the other handle in the other half of S . WL does not embed in S but its boundary 3-manifold does.

25 (Note that WL with one handle surgered to be a 1-handle does embed; i.e. slice knot ∪ dotted unknotted circle. This is because the 1-handle denotes something to be scooped out of the ﬁrst B4, and the 2-handle is then attached into the other half.)

4 ∼ Next, we must consider 4-manifolds with 1-handles, because π1(B \ slice disc) = Z. These can still be drawn with link pictures, where we use dotted circles to indicate 1-handles.

Theorem 3.22 (Laudenbach). A closed 4-manifold N 4 with a handle decomposition is determined by its 2-skeleton. This follows from the fact that any diﬀeomorphism of ]S1 × S2 extends over ]S1 × B3.

Remark 3.23. His proof is by the classiﬁcation of the mapping class group of ]S1 × S2.A particular element is the Gl¨uck twist:

S1 × S2 → S1 × S2

(x, v) 7→ (x, x.v) x ∈ SO(2) ⊂ SO(3) which acts on S2 We can cut out S2 × B2 from a 4-manifold and reglue it in this way, which might be a way to get an exotic smooth structure on S4.

4 Back to Classical Knot Concordance

In this section we have knots S1 ⊆ S3, which are slice if they bound properly embedded, locally ﬂat discs D2 ⊆ D4, such that ∂D2 = S1, ∂D4 = S3.

4.1 The Cappell-Shaneson way to slice a knot 3 Lemma 4.1. If K is slice then MK = 0-framed surgery on K bounds a 4-manifold W such that ∼ ∼ (i) H1(M) = H1(W ) = Z with the isomorphism induced by the inclusion M ⊆ W ;

(ii) H2(W ) = 0;

(iii) π1(W ) is normally generated by the meridian of the knot. Theorem 4.2 (Freedman). The converse holds in the topological locally ﬂat category.

Remark 4.3. If (i) and (ii) are satisﬁed then K is slice in a homology 4-ball. We call such a knot homologically slice. However, there are no known obstructions which are able to tell the diﬀerence between homologically slice and actually slice.

26 Proof of Lemma 4.1. Deﬁne W 4 = B4 \ (D2 × D2), a 4-ball minus a thickened slice disk for the knot. Then ∂W = MK . It is the 0-framed surgery because the push-oﬀ of K along the slice disk, towards its centre, does not link with K. Then (i) and (ii) follow from Alexander duality, or a 2 1 3 1 2 Mayer-Vietoris calculation. Decomposing MK = X ∪(S1×S1) (D × S ), where X = S \ (S × D ) ∼ is the knot exterior, the Mayer-Vietoris sequence yields that Hi(MK ; Z) = Z for i = 0, 1, 2, 3, and 4 2 2 is of course zero otherwise. Decomposing B = W ∪D2×S1 (D × D ), Mayer-Vietoris implies (i) and (ii). (iii) follows from the geometric fact that adding a handle which kills the meridian makes the group 4 2 2 zero. This can be seen with the Seifert-Van Kampen theorem: since D = W ∪D2×S1 D × D . The boundary of D2 × D2 is D2 × S1 ∪ S1 × D2, but the S1 × D2 represents the neighbourhood of the knot: W and D2 × D2 are just glued together along D2 × S1 which is properly embedded in D4. Therefore: 2 2 4 ∼ ∼ π1(W ) ∗ π1(D × D ) ∼ π1(W ) π1(D ) = 1 = 2 1 = 2 1 π1(D × S ) π1(D × S ) 4 which shows that the image of the meridian generates π1(W ) as claimed. If B were just a homology 4-ball, then (i) and (ii) still hold, but (iii) might not.

Proof of Remark 4.3. Suppose there is a W such that (i) and (ii) hold. Start with this W , and 4 attach a 2-handle to W along the meridian - do the reverse surgery to that which made MK from 3 2 2 3 3 S . Then W ∪MK (D × D ) is a homology 4-ball. Its boundary is S , and the knot lives in S as the belt sphere of the attached 2-handle {0} × S1 ⊆ D2 × D2, and the cocore of the handle is a homology slice disk for the knot.

Proof of Theorem 4.2. Start as above, making a homology 4-ball B with a slice disk. If we also have condition (iii), then π1(B) = 0 so we have a homotopy 4-ball. We can then apply Freedman’s

27 topological Poincaréhypothesis show that B is homeomorphic to B4. This is a strange homeomorphism: the same strangeness which means that it may not be smoothly a 4-ball, and which made embedding Casson handles so hard. The slice disk looks strange but it is locally flat because the homeomorphism carries a regular neighbourhood with it. The smooth case of this is an open question: is a homotopy 4-ball diffeomorphic to B4?

Remark 4.4. There are examples of knots which are topologically locally-ﬂat but not smoothly slice: for example, the Whitehead double of a 0-framed trefoil had Alexander Polynomial 1 and so is topologically slice, however results of gauge theory which estimate the slice genus can show that it is not smoothly slice. The Cappell-Shaneson way to slice a knot is, instead of looking for the slice disk, to start with a slice disk in a 4-manifold with boundary MK and try to change the 4-manifold so that the conditions of Lemma 4.1 are satisﬁed. In this way the machinery of surgery theory can be employed to build an obstruction theory.

4.2 Drawing 4-manifolds with 0, 1, and 2-handles We have already discussed the case where there are no 1-handles. To attach a 1-handle to B4, we attach a copy of B1 × B3 to a pair of embedded 3-balls in the boundary S3. Since 1- and 2-handle pairs cancel, this is the same as removing a 2-handle from the interior of B4 (it is helpful to ﬁrst imagine the picture in 1 dimension fewer, and then cross everything with I). For all the 1-handles to be attached, we indicate the belt spheres {0} × S1 of the 2-handles to be removed from the interior by drawing unknotted, unlinked dotted circles in S3 (in the 3-dimensional case we would indicate an S0 in the boundary S2). The 1-handles generate the fundamental group. We can then draw a framed link in S3, which may link with the dotted circles, to indicate the attaching spheres for the 2-handles with their framing.

Theorem 4.5. The diﬀeomorphism type of WL is unchanged under the following moves on L: (a) Isotopy of L.

(b) Handle slides of one handle over another. This means taking connected sum of one component of L with a push oﬀ of another, using the framing to perform the push oﬀ (1-handles have zero framing for this purpose). This is legal for 1-handles on 1 handles, 2 on 1, and 2 on 2.

(c) After ﬁrst pushing them both oﬀ everything else, cancellation of a 1- and 2-handle pair. They will be represented in L by a Hopf link in which one of the components is dotted.

Remark 4.6. We can compute π1(WL) by writing relations corresponding to the way in which the 2-handle attaching maps link the dotted circles, which represent generators of the fundamental group. We can compute H∗(WL) using the handle chain complex. Actually, we can compute the handle chain complex and homology of the universal cover of W : replace each Z in the chain complex with a copy of Z[π1(W )], and read oﬀ the intersection matrix with Z[π1(W )] coeﬃcients.

28 These can be deﬁned if each of the handles is attached to a base point with some choice of path; each intersection point between two handles comes with an element of π1(W ) given by concatenating the paths belonging to each of the handles. π2(WL) = π2(WgL) = H2(WgL; Z) = H2(WL; Z[π1(W )]) can be computed using the handle chain complex as described. Remark 4.7. When sliding handles over one another, we cannot do a band sum going through the spanning discs for the 1-handles, since the spanning disk has been removed from B4; recall that this is what the dotted circle denotes.

Remark 4.8. If L = L1 q L2, then WL = WL1 ]WL2 ; ∂WL = ∂WL1 ]∂WL2 . This means that 2 adding a ±1-framed unknot to L does not change the boundary of WL: it changes WL to WL]CP . This is called blowing up, and the reverse procedure is called blowing down. The name comes from algebraic geometry, where the technique is used to blow up unwanted singularities of complex 2 surfaces by adding the freedom of an extra copy of CP . (See Gompf and Stipsicz, Section 2.2) Theorem 4.9 (Kirby). Two 3-manifolds M 3 ∼ M 3 are diffeomorphic if and only if L can be L1 = L2 2 obtained from L1 by a finite sequence of isotopies, handle slides, births/deaths, and blowing up and down. Remark 4.10. For any smooth manifold, handle decompositions exist, since there exists a Morse function and a gradient-like vector field on the tangent space. They are unique up to handle slides and cancellation (Cerf theory - it has to be shown that the handles can always be arranged in ascending or descending index with respect to the Morse function.)

4.3 Which 4-Manifolds does MK bound? First attempt: 4 MK = ∂(B ∪K 2-handle) The trouble with this is that it is simply connected but has second homology non-zero. We want to get one with fundamental group equal to Z, so that the first homology is also Z. 4 i∗ ∼ Theorem 4.11. (a) There always exists a W with ∂W = MK and H1(M) −→ H1(W ) = Z an ∼ isomorphism. (and π1(W ) = Z). The H2(W ) = 0 condition of Lemma 4.1 then gives rise to the obstruction to slicing the knot. π2(W ) = H2(Wf) is a free Z[Z]-module of the same rank as the rank of H2(W ) over Z. (b) Let λ = I be the intersection form on the universal cover of W . Then representing the Wf −1 intersection form as a matrix over Z[π1(W )] = Z[Z] = Z[m.m ] we have that det(λ)(m) = ∆K (m), the Alexander polynomial of K. (c) Our W 4 from (a) has a twisted signature σ(I ⊗ (z)), which coincide with the original Wf C 1 definition for a knot signature σz(K), for z ∈ S ⊆ C. C(z) is the field of fractions (so that a signature is well-defined) of the ring of polynomials in z with complex coefficients, and z is a ∼ n 1 character on Z = π1(W ), n 7→ z ∈ S ⊂ C.

29 Remark 4.12. Recall that if a knot K is slice then it is homologically slice, so MK is the boundary ∼ of N with H1(N) = H1(MK ),H2(N) = 0. This last condition, however, would force the intersection form to be trivial and hence the Alexander polynomial to be 1. Since such knots are already known to be slice, we are interested in those without this property, and so will need 4-manifolds with more complicated groups than Z in order to apply the Cappell-Shaneson method to the classical dimension.

Proof of Theorem 4.11. (a) Pick a Seifert surface F for the knot, and draw it in band form. Note that the linking and framing of the bands deﬁne the Seifert matrix of linking numbers, which will be of the form AC V = , I + CT B where A = AT , B = BT , and V − V T is symplectic. We can use this presentation of the Seifert surface to draw a link in S3, and thus to deﬁne a 4-manifold with the desired properties. The central disc of the Seifert surface becomes a dotted circle, and the bands become 2-handle attaching maps, whose embeddings depend on the knotting and linking of the bands. Figure 7 shows the procedure.

“a” curves, with framing according to twists in the corresponding bands “b” curves, 0-framed but with correct number of twists.

2 0

Repeated accordingly for the genus of the Seifert surface

Linking 1 here, corresponding to symplectic

intersection form of H1(F) Dotted circle = 1-handle Overall linking of 2-handle attaching maps with 1-handle circle is zero.

∼ Figure 7: Drawing a 4-manifold W0 with π1(W0) = Z and ∂W0 = MK .

30 ∼ So W0 is comprised of a 0-handle, a 1-handle, and 2g 2-handles. π1(W0) = Z since there is a single 1 W 2 1-handle which has zero algebraic linking numbers with each of the 1-handles. W0 ' S ∨ 2g S . ∼ ∼ L ∼ L ∼ 2g Thus π2(W0) = H2(Wf0) = 2g Z[π1(W0)] = 2g Z[Z], while H2(W0) = Z . To see that this 4-manifold has boundary MK , we have to apply some Kirby moves to change this link into the knot: change the 1-handle to a 2-handle and slide it twice over each of the “a”-curves, using a band which follows the “b”-curves to perform the connected sum for one of the slides, and the obvious small band for the other one. This leaves the a and b curves as geometric duals which can be cancelled, and leaves the old 1-handle zero framed and in the shape of K. ∼ (b) To compute the intersection form λW on H2(Wf) = π2(W ), we take a basis of immersed 2 spheres fi : S # W , with a threading. A threading is a path from the basepoint of W to the 2 basepoint of fi(s) where s is a basepoint of S . Such spheres arise as the unions of the cores of the 2-handles with immersed discs which form null-homotopies of the attaching maps of the 2-handles inside S1 × B3, which is the 4-manifold created by attaching a single 1-handle to B4. These null- homotopies exist because each of the 2-handle attaching maps link the dotted circle zero times. We count intersections of our spheres X (fi, fj) = ε(x) · g(x) 2 2 x∈fi(S )∩fj (S ) where ε(x) is the sign which arises from comparing orientations of the tangent bundles at a transverse intersection, and g(x) is an element of π1(W ) which is the composite of the threading of fi, paths from fi(s) to x and then from x to fj(s), followed by the inverse of the threading for fj. The self intersections are counted by taking a parallel copy of each fi and counting as usual. (Not to be confused with the quadratic self-intersection µ-form which we have not introduced.) Note that we must have spheres and not surfaces (we could guarantee embedded surfaces to represent each homology class of W ), unless there are conditions on the fundamental groups of the surfaces, since otherwise they will not lift to the covering space due to their non-trivial fundamental groups. The intersection form satisﬁes λ(f1, f2) = λ(f2, f1) −1 where the involution used is induced from g 7→ g on π1(W0). There is no sign change when switching variables since the dimensions are even. In order to relate the intersection form to the Alexander polynomial, consider the long exact sequence of the pair (W0, ∂W0 = MK ) with Z[Z] coeﬃcients:

λ H2(W0; Z[Z]) −→ H2(W0, ∂W0; Z[Z]) → H1(MK ; Z[Z]) → H1(W0; Z[Z]) ∼ ∼ π1(W0) = Z, so H1(W0; Z[Z]) = 0. Poincaré-Lefschetz duality and universal coefficients mean that H (W , ∂W ; [ ]) ∼ H2(W ; [ ]) ∼ Hom (H (W ; [ ]), [ ]). This justifies the labelling 2 0 0 Z Z = 0 Z Z = Z[Z] 2 0 Z Z Z Z of the arrow λ; it is indeed the intersection form. H2(W0; Z[Z]) is free since there are no 3- ∼ handles of W0. We therefore have H1(MK ; Z[Z]) = coker(λ), and since the Alexander polynomial is the characteristic polynomial of this Z[Z]-module, we have that ∆K = det(λ). Note that the

31 determinant of λ depends on a choice of basis, which could vary by units in Z[Z]; however the Alexander polynomial is also only ever defined up to a unit. (c) Now, to compute the signature of the intersection form, we give an explicit matrix for it, using the explicit 4-manifold W0 depicted in Figure 7. The immersed 2-spheres which form our basis of H2(W0; Z[Z]) are given, as usual, by the cores of the 2-handles combined with immersed discs in S1 × B3. These discs will have self intersections governed by the framing of the link components, and will intersect one another depending on the way the components link one another. We count these intersections with coefficients in hmi governed by linking with the dotted circle which denotes the 1-handle. We claim that: AI + (m − 1)C λ = −1 T −1 Wf0 I + (m − 1)C (m − 1)(m − 1)B where each of A, B, C are block g × g matrices from above, so A = AT and B = BT . For the off diagonal entries, we try to homotope the a curves away from the b curves (look again at Figure 7). Since the b curves imitate the bands of a Seifert surface, we get pairs of intersection points, and an extra intersection at the end (the +1) due to the single linking of the a and b curves. Since the two strands of the b curves differ by a single passage around the 1-handle i.e. through the dotted circle, this explains the (m − 1) coefficient of C in the top right. The Hermitian property of λ fills in the bottom left square for us. For the b-b pairs, each linking of the bands induces four intersection points: hence (m − 1)(m−1 − 1)B. Similarly λ-self intersections, that is homological self intersections, so intersections with a pushed off copy of the sphere, produces 4 intersection points for each twist in the band. Note here the difference with the µ-form which counts actual self intersection points. The intersections of the a curve spheres depend on the framing of the a curves which were taken from the linking of the corresponding bands in the Seifert surface and do not depend on m. Now, tensor with C(z), so that we can invert polynomials. We have AI + (z − 1)C λ = −1 T −1 Wf0 I + (z − 1)C (z − 1)(z − 1)B

Make a change of basis with change of basis matrix

(z − 1)I 0 P = 0 I

In particular note that z − 1 can be inverted now, so P is an invertible matrix. Also, in the calculation below, the key step is that q = z − 1 is a quadratic element, so that qq = q + q. The

32 matrix of λW0 becomes: T AI + (z − 1)C λ = P P W0 I + (z−1 − 1)CT (z − 1)(z−1 − 1)B (z−1 − 1)I 0 AI + (z − 1)C (z − 1)I 0 = 0 I I + (z−1 − 1)CT (z − 1)(z−1 − 1)B 0 I (z − 1)(z−1 − 1)A (z−1 − 1)I + (z − 1)(z−1 − 1)C = (z − 1)I + (z − 1)(z−1 − 1)CT (z − 1)(z−1)B ((z − 1) + (z−1 − 1))A (z−1 − 1)I + ((z − 1) + (z−1 − 1))C = (z − 1)I + ((z − 1) + (z−1 − 1))CT ((z − 1) + (z−1))B AC AT C + I = (z − 1) + (z−1 − 1) since A = AT and B = BT I + CT B CT BT = (z − 1)V + (z−1 − 1)V T the signature of which was deﬁned earlier to be the signature of the knot σz(K). If z = 1, this AI change of basis is not possible; however, we then have σ (K) = 0, and the matrix , so 1 I 0 we still have agreement.

Example 4.13. Consider the link shown in Figure 8. This is a Kirby diagram for a 4-manifold ∼ ∼ ∼ ∼ −1 W with π1(W ) = Z = hmi, π2(W ) = Z[π1(W )] = Z[m, m ] (since W has just one 2 handle it 1 2 ∼ is homotopy equivalent to S ∨ S ) and H2(W ) = Z. The Hurewicz homomorphism is given by augmentation m 7→ 1. The intersection form IW = (1), as given by the framing on the 2-handle attaching map.

4 ∼ Figure 8: The Whitehead link with one component dotted as an example of W with π1(W ) = Z and ∂W = MK .

33 We can also compute the intersection form I of the universal cover, which here is the infinite Wf ∼ −1 cyclic cover. We can generate the second homology H2(Wf; Z) = H2(W ; Z[m, m ]) with an immersed 2-sphere constructed from the union of the core of the 2-handle with a disc in S1 × B3 with one self intersection. We cannot use the core union a Seifert surface here because the Seifert surface isn’t a disc, and so doesn’t extend to the universal cover as a surface because it has non-trivial fundamental group. Passing through the dotted circle takes us to a different sheet of the covering. The attaching circle of the 2-handle links adjacent copies of itself with linking number −1 and links itself with linking number 1. This corresponds to analogous intersections of the corresponding 2-spheres. Therefore I = (−m − m−1 + 1) Wf which is the Alexander polynomial of the trefoil. Furthermore, we can work out the boundary of W as follows. First, change the 1-handle to a zero framed 2-handle. Then exploit the symmetry of the Whitehead link to obtain the same picture as in Figure 8 but with the dot replaced with a +1 and the +1 replaced with a 0 (See Figure 9, where the link components are labelled x and y). Slide the handle labelled x twice over the handle labelled y, once for each strand of x which passes through y. This should be done with orientations opposite on each handle slide. The effect of this is to move y away from x, and to introduce a full twist in x, without changing the framing. We can then blow down y to leave a trefoil knot with 0 framing. Therefore W has boundary M31 , the zero-surgery on the trefoil.

+1 x

Figure 9: Finding the boundary of W .

5 L2-homology

5.1 Introduction So far we do not have any new concordance invariants beyond the “high-dimensional” invariants which detect non-algebraically slice knots: we have only looked at 4-manifolds with fundamental

34 group Z. In order to obtain more sophisticated invariants we need to look at 4-manifolds with larger groups, and their equivariant intersection forms. The group rings will be non-commutative rings; our tools (to be defined below) for detecting the signatures of forms over non-commutative rings will be the L2-homology and the L2-signature. We need a generalisation of the idea of dimension for submodules after embedding the group ring in its Von Neumann algebra. This enables us to look at the difference between the dimensions of the positive and negative eigenspaces of a Hermitian operator on a Hilbert space corresponding to a symmetric sesquilinear form, and so to define a real number which is the L2-signature, giving a homomorphism from the Witt group of such forms 2 to R. When defining this for a knot, to get a concordance invariant, we look at the L -signature of the equivariant intersection form of a 4-manifold whose boundary is MK . We then define the reduced L2-signature which is an invariant of the zero surgery, so does not depend on the choice of 4-manifold. In order to calculate these signatures we are therefore able to make judicious choices of 4-manifold, whose intersection forms we can exhibit explicitly. In the case of W0 above, it turns out that the L2-signature is given by integrating the twisted signatures of the knot around S1.

5.2 Atiyah’s original definition of L2 Betti Numbers Suppose M is a compact Riemannian manifold, and M → M is a regular cover with an infinite group as deck transformations, so M is non-compact. The idea is to measure the space of smooth L2-integrable p-forms on M (if they exist - this is not immediately clear). It will be either zero or infinite dimensional, so we want a measurement of dimension which is zero if and only if there are no such forms. Thus, the analytic L2-Betti number is Z (2) t4p bp (M) = lim trC(e (x, x)) dx t→∞ a fund. domain for M

p p • 4p :Ω (M) → Ω (M) is the Laplacian on p-forms (lift the metric).

• et4p (x, x) is the heat kernel.

p • trC means the normal matrix trace of the section of End(Λ (T M)) i.e. ﬁbre-wise trace. • The group Γ of the covering is brought in by the choice of fundamental domain i.e. it cannot just be an invariant of the covering.

Remark 5.1. These Betti numbers are zero for circle and tori, but are non-zero for surfaces and 2 H covering spaces.

5.3 Alternative Approach

If Xe → X is the universal covering of a ﬁnite CW-complex X, and M is a π1(X) module, then we can deﬁne H (X; M) = H (C (X) ⊗ M). p p ∗ e Z[π1(X)]

35 For example, suppose we have a surjection π1(X) Γ. Then we can define Hp(X; Z[Γ]) = H (C (X) ⊗ [Γ]) = H (X), the homology of the cover corresponding to Γ. H (X) is p ∗ e Z[π1(X)] Z p p a Z[Γ] module, but algebraically it is not at all well understood - it might not even be finitely generated, since the ring Z[Γ] is non-Noetherian which means that submodules of finitely generated modules may not be finitely generated. The idea is then to embed Z[Γ] ,→ N Γ, the Von Neumann Algebra of Γ, a ring with a much simpler representation theory. Then:

Definition 5.2. 2 Hp(X) = Hp(X; N Γ) is the L homology; and (2) bp (X) = dimN Γ Hp(X; N Γ) ∈ [0, ∞) . We now define the Von Neumann Algebra. Let Γ be a countable discrete group and consider the Hilbert space l2Γ of square-summable sequences of group elements with complex coefficients.

2 {Σagg | ag ∈ C, Σ|ag| < ∞}. The group Γ acts by left- and right- multiplication on l2Γ. These operators are obviously isometries and we can consider the embedding 2 CΓ ,→ B(l Γ) corresponding to (sums of) right multiplications into the space of bounded operators on l2Γ.

∗ ∗ Deﬁnition 5.3. The (reduced) C -algebra C Γ is the completion of CΓ with respect to the operator 2 norm on B(l Γ). The von Neumann algebra N Γ is the completion of CΓ with respect to pointwise 2 ∗ convergence in B(l Γ). In particular we have CΓ ⊆ C Γ ⊆ N Γ. Von Neumann’s double commutant theorem says that N Γ is equal to the set of bounded operators which commute with the left Γ-action on l2Γ:

N Γ = B(l2Γ)Γ

Example 5.4. (a) |Γ| < ∞ ⇒ N Γ = C[Γ] which is semisimple, so the algebra is manageable. ∞ 1 1 (b) Γ = Z: then N Γ = L (S ; C), the space of (almost everywhere) bounded functions S → C. C[Γ] sits inside this as Laurent polynomials.

5.4 Trace We deﬁne a trace operator in order to measure dimension of subspaces: recall that if we have a subspace U of a ﬁnite dimensional vector space V , then choosing a basis and representing the projection pr : V → U as a matrix, and taking its trace tells us the dimension of U; this follows easily from the fact that the trace is invariant under change of basis, and so in a sense the trace

36 tells us the deﬁnition of the projection operator. We need to generalise this to the setting of Hilbert spaces operators and thus to Von Neumann algebras; we now get a real valued idea of dimension. We deﬁne:

tr: N Γ → C by:

a 7→ ha(e), eil2Γ, 2 where e ∈ l (Γ) is the identity element, so that for a ∈ CΓ, tr(a) = ae = coeﬃcient of e. (This is not the usual augmentation.) It satisﬁes:

(i) tr(ab) = tr(ba). To see this, it is enough to show that it is true for a ∈ N Γ and g ∈ Γ by linearity. But hag(e), ei = h(ae)g, ei = hae, g−1ei = hg(ae), ei using the bi-module property and that the left and right Γ-actions are isometries.

(ii) tr is faithful, i.e. tr(a∗a) = 0 if and only if a = 0. This is because ha∗a(e), ei = hae, aei = 0 ⇔ ae = 0.

But then ag = 0 for all g ∈ Γ by right multiplication and then by continuity a · CΓ = 0 so a = 0.

Deﬁnition 5.5. Let P be a f.g. projective N Γ module; i.e. P = im(p) for some p ∈ Mn(N Γ) with p2 = p = p∗. Then we can use the trace operator to deﬁne the Γ-dimension:

dimN Γ P = trΓ(p), where the right hand side is given by the sum of the Γ-traces of the diagonal entries of p. Since p = p∗ and tr(a∗) = tr(a), this is a real number, in fact a positive real number: tr(p) ∈ [0, ∞) since: tr(p) = tr(p2) = tr(p∗p) = hpe, pei ≥ 0. We therefore have a map: K0(N Γ) → R. Theorem 5.6 (Farber,L¨uck). N Γ is a semi-hereditary ring: any f.g. submodule of a projective module is projective.

Corollary 5.7. Given X → X, there are ﬁnitely generated projective modules Pi,Qi such that the following is exact: (2) 0 → Pi → Qi → Hi (X) → 0 2 (2) (The L homology modules have homological dimension 1.) Thus we can deﬁne, for Hi (X), which is not necessarily projective, the Betti numbers: (2) bi (X) = dimN Γ Qi − dimN Γ Pi.

37 Remark 5.8. According to W. L¨uck we can actually extend the dimension to all modules over N Γ.

2 2 1 Remark 5.9. Specialise again to Γ = Z: l (Z) is the Fourier expansion L (S ; C) but taking ∞ 1 closure in bounded operators gets us as above to L (S ; C) which is all bounded functions modulo diﬀerences on sets of measure zero; trace becomes integration over S1, since in hf.e, ei, e is the constant function 1 and inner product is integral.

Let A be a C∗ algebra with involution ∗. Pick a ∈ A which is normal, that is aa∗ = a∗a, and consider spec(a). Since A is a C∗-algebra, spec(a) is compact, so there is a ∗-ring homomorphism.

0 C (spec(a); C) → A; f 7→ f(a)

1 which sends Idspec(a) 7→ a, and the constant 1 7→ 1. We send functions to S to units, polynomials to polynomials and real valued functions to hermitian operators. a generates an abelian subalgebra which equals C0(spec(a)) by Gelfand-Naimark. If A is a Von-Neumann Algebra, then the above ap- ∞ plies, and moreover it also applies with L (spec(a); C) → A, i.e. with bounded almost everywhere functions rather than just with continuous functions. This map is the functional calculus - starting with a normal element of our Von Neumann algebra we obtain a new element for each function on the spectrum. For A = Mn(N Γ), this element can be used to define projective submodules of A. ∗ 2 (2) Given h ∈ Mn(N Γ) which satisfies h = h , we want to define the L -signature σΓ (h) ∈ R. 2 To define the L -signature, consider a hermitian (n × n)-matrix over N Γ, h ∈ Hermn(N Γ), as a bounded, hermitian Γ-equivariant operator on the Hilbert space (l2Γ)n. Its spectrum spec(h) is a (compact) subset of the real line and for any bounded measurable function f on spec(h) we may define the bounded Γ-equivariant operator f(h) ∈ Mn(N Γ) by functional calculus. In particular, consider the characteristic functions p+, p−, p0 : R → R of (0, +∞) respectively (−∞, 0), {0}.

Theorem 5.10 (Functional Calculus). There are ﬁnitely generated projective N Γ-modules H+,H−,H0 such that after a base change (h 7→ aha∗), there is an orthogonal decomposition

2 n H := (l Γ) = H0 ⊕ H+ ⊕ H− such that  0  h =  +1  −1

2 n Proof. h is a Hermitian operator in H := B((l Γ) ). Then spec(h), the set of λ ∈ C such that h − λ.Id is not invertible, is a compact subset of R. (Bounded operators have compact spectra; Hermitian operators have real elements in their spectra.) Then, given any bounded function f : spec(h) → R, there is a Hermitian operator f(h) with spec(f(h)) = f(spec(h)). The proof of this is that for polynomial f, f(h) is deﬁned uniquely. In

38 general, f is the point-wise limit of polynomials fi: define f(h) = point-wise limit of fi(h), which by definition lives in Mn(N Γ). Note that, as above, the functional calculus ∞ L (spec h, C) → Mn(N Γ); f 7→ f(h) ∗ is a C -algebra homomorphism. Now define f+, f−, f0 by Heaviside type functions, so that we use the characteristic functions of the intervals (0, ∞), (−∞, 0) and the singleton {0} in R. 1 = f+ + f− + f0, so h = f+(h) + f−(h) + f0(h) giving an orthogonal decomposition of the Hilbert space as H+ := f+(h)(H), H− := f−(h)(H) and H0 := f0(h)(H). For example, f+(h) ∈ Mn(N Γ), so its image is a projective submodule of H. Since the functional calculus is a homomorphism, the maps f+, f−, f0 are commuting projections, and so we get an orthogonal decomposition as claimed.

Definition 5.11. The signature map σΓ : Hermn(N Γ) → K0(N Γ) is defined by sending h to the 2 formal difference p+(h) − p−(h) of projections in Mn(N Γ). The L -signature of h ∈ Hermn(N Γ) is defined to be the real number (2) σΓ (h) := trΓ(p+(h)) − trΓ(p−(h)) = dimN Γ P+ − dimN Γ P−.

(2) 2 ∗ Remark 5.12. |σΓ (h)| < n. To see this note that dimN Γ(N Γ) = 1; since p = p = p acts on one N Γ. Also 0 ≤ tr(p) ≤ 1: just use (1 − p)∗ = (1 − p) as well, so tr p ≥ 0, tr(1 − p) ≥ 0. 2 Lemma 5.13. The L -signature only depends on the Γ-isometry class of h ∈ Hermn(N Γ), i.e. it ∗ is unchanged under h 7→ a ha for a ∈ GLn(N Γ). Proof. Consider the Hilbert space H := (l2Γ)n with the bounded Γ-equivariant operators h and a. We have an orthogonal decomposition of Hilbert spaces

H = H0 ⊥ H+ ⊥ H− as above. For a vector v in one of the three summands above, one has by deﬁnition that hh(v), vi = 0, > 0 respectively < 0.

−1 −1 −1 It follows that depending on whether v is in a H0, a H+ respectively a H− one has h(a∗ha)(v), vi = hh(av), avi = 0, > 0 respectively < 0. Therefore, the three orthogonal projections

−1 ∗ a H† → p†(a ha)H for † ∈ {0, +, −} are monomorphisms and thus

−1 ∗ dimΓ H† = dimΓ a H† ≤ dimΓ p†(a ha)H for † ∈ {0, +, −}. But the three dimensions on both sides must sum up to the total dimension n of H and therefore the inequalities are actually equalities.

39 Example 5.14. We recapitulate the crucial example, and consider once more the case Γ = hti =∼ −1 Z. Then CΓ consists of Laurent polynomials C[t, t ] which embed naturally into the space of complex valued continuous functions on the circle S1. Indeed, Fourier transformation gives an 2 ∼ 2 1 isomorphism of Hilbert-spaces l Z = L (S ; C) and point-wise multiplication by a function induces ∗ ∼ 1 the isomorphism C Γ = C(S ; C). This is a consequence of the Stone-Weierstraß theorem on the density of polynomials in the space of all continuous functions in the supremum norm. Completing in the topology of pointwise convergence then leads to the von Neumann algebra N Γ which turns ∞ 1 out to be the space L (S ; C) of complex valued, bounded, measurable functions on the circle, deﬁned almost everywhere. Finally, the standard Γ-trace is just given by integration. 1 1 Now consider h ∈ Hermn(C(S ; C)) and think of it as a continuous map from S to Hermn(C). The ordinary signature σ : Hermn(C) −→ Z counts the number of positive Eigenvalues minus the number of negative Eigenvalues.

1 Deﬁnition 5.15. The twisted signature of h is the step function σ(h): S −→ Z which assigns to each s ∈ S1 the signature σ(h(s)). Moreover, the real number σ(2)(h) is deﬁned to be the integral of this function σ(h) over the circle (normalized to have total measure 1).

Thus σ(2)(h) is the average of all the twisted signatures. It is clear that σ(h) makes sense almost ∞ 1 (2) everywhere for h ∈ Hermn(L (S ; C)) and therefore σ (h) is well deﬁned even in this case. As ±1 an example, consider the following element in Herm2(C[t ]): t + t−1 − 2 t − 1 h := t−1 − 1 t + t−1 − 2

Notice that σ(h) is a step function with jumps at most at the zeroes of the “Alexander polynomial” det(h) ∈ C[t, t−1]. We have, up to multiplication by a unit ±tn, det(h) = (t + t−1 − 2)2 − (t − −1 2 −1 2 genus(31) 1 1)(t − 1) = (t − 1) (t + t − 1) = (t − 1) ∆31 (t) which has roots on S exactly for the two primitive 6-th roots of unity ±eiπ/3, and for t = 1. For t = 1 we have the zero matrix and so this does not contribute to the signature. We only need to calculate√ σ(h) at two points on the circle which interlace with the two roots of ∆ (t), e.g. at −1 and 3/2 + i/2. One easily checks 31 √ that the ordinary signature of h(−1) is −2, and of h( 3/2 + i/2) is zero. Since these roots are of distance 1/3 and 2/3 apart, depending on which way round of the circle we measure, we therefore have: σ(2)(h) = (1/3) · 0 + (2/3) · (−2) = −4/3 6= 0. Of course this Hermitian operator is equivalent to the equivariant intersection form of our 4- manifold W0 from Theorem 4.11 for the trefoil knot, so has the twisted signatures of the trefoil for 1 each s ∈ S ⊆ C. It is possible to similarly work this out for all twist knots and for torus knots (recall that the torus knots arise as the self-linking zero curves on the obvious Seifert Surfaces for the twist knots).

(2) 2 (2) Lemma 5.16. The average σ (h) equals the L -signature σΓ (h) for Γ = Z.

40 Proof. Notice that trΓ(p+(h)) is the Γ-dimension of the “positive Eigenspace” of h. In the functional calculus one approximates (say) p+ by a sequence of real polynomials pi into which any operator can easily be substituted. Then one takes the pointwise limit to deﬁne

p+(h) := lim pi(h) for h ∈ N Z. i

For example, if h is a finite dimensional matrix, then one checks that p+ is just the projection onto the (+1)-Eigenspace of h. This implies that for a point s ∈ S1, a fancy way to count the number of positive Eigenvalues of h(s) ∈ Hermn(C) is to take the ordinary trace of p+(h(s)) := limi pi(h(s)). But now one clearly sees that the integral of the function σ(h) which associates for each s ∈ S1 the (2) difference p+(h(s)) − p−(h(s)) is almost everywhere the same as σΓ (h). Corollary 5.17. Z σ(2)(h) = σ(h(z)) dz Z S1 ∗ 2 Remark 5.18. Note that the orthogonal projections (p = p = p ) in N Z correspond one to one with measurable sets in S1. ∗ ∞ 1 p = p ⇒ p ∈ L (S , R); p2 = p ⇒ p has values 0 or 1 almost everywhere, so we can view it as a characteristic function. A consequence of this is that any positive real number occurs as a dimension of a projective module over N Z: just use the function p to define pN Z with dimension p, 0 ≤ p ≤ 1. Then add copies of N Z to make larger numbers. ∼ ∞ 1 Also K0(N Z) = L (S ; Z). This follows since a projection operator can be diagonalised without ∗ 2 changing the trace, with each diagonal element satisfying pi = pi = pi as above, which means it picks out a measurable subset of S1 as the characteristic function. Taking the matrix trace of the 1 trΓ’s yields the sum of these characteristic functions which is a bounded measurable function S → Z. Negative numbers are associated to the projective modules in K0(N Z) which are included as formal inverses in the Grothendieck completion which defines the projective class group. Integrating ∞ 1 the elements of L (S ; Z) yields an epimorphism to R.

5.5 Properties of L2 Betti Numbers

(2) 2 Γright Recall Hi (X) = Hi(X; N Γ), where ZΓ ⊆ CΓ ⊆ N Γ = B(l Γ) , which is a completion of CΓ under point-wise convergence of operators. We can actually think of the chain complex:

] i cells Ci(X) ⊗ZΓ N Γ = (N Γ) with the boundary maps being matrices di ∈ Mn(N Γ).

41 Property 1. Homotopy Invariance If f : X ' Y is a homotopy equivalence which lifts to a (2) (2) homotopy equivalence of the Γ-cover, then Hi (X) = Hi (Y ). (This would be surprising using the analytic integral of the heat kernel deﬁnition of L2-Betti numbers.) Recall that Hi(X) is not quite a projective module: it can be given by

(2) 0 → Pi → Qi → Hi (X) → 0

(2) where Hi (X) decomposes as a f.g. projective part, as measured by the dimension, and a torsion part, as measured by the Novikov-Shubin invariants.

Property 2. Euler-Poincar´eFormula For a ﬁnite complex:

X i X i (2) χ(X) = (−1) bi(X) = (−1) bi (X)

Property 3. Multiplicative under ﬁnite covers Consider the tower of covering spaces

Γ0 X / X0 }} Γ }} }} n ~}} X where Γ0 has index n in Γ. Then

(2) 0 (2) bi (X )Γ0 = n · bi (X)Γ. Proof of Property 2.

X i X i X i X i (2) (−1) bi = (−1) dimQ(Ci(X) ⊗ Q) = (−1) dimN Γ(Ci(X) ⊗ N Γ) = (−1) bi (X) just from additivity of rational dimension under short exact sequences and similarly for the Von Neumann dimension. Both chain complexes consist of free modules whose rank in each dimension is equal to the number of cells of that dimension. Since X is a ﬁnite complex, this is legal.

Proof of Property 3. The algebraic statement necessary is that Γ0 ≤ Γ induces N Γ0 → N Γ; if M is an N Γ-module then we can restrict to N Γ0 with dimension given by

dimN Γ0 (resN Γ0 M) = n · dimN Γ M.

(2) Conjecture 5.19 (Atiyah Conjecture). Γ torsion free implies that bi (X)Γ ∈ N. This is known for elementary amenable groups, free groups, class closed under certain operations (not hyperbolic yet).

42 Property 4. Poincar´eDuality If X is a closed oriented manifold of dimension n, then (2) (2) bi (X)Γ = bn−i(X)Γ. The proof is that the usual chain equivalence between homology and cohomology complexes, tensored with N Γ, remains a chain equivalence. There is a subtlety, namely that we have to use L2-cohomology, but it is very close to L2-homology.

Property 5. K¨unneth formula

(2) X (2) (2) bn (X1 × X2)Γ1×Γ2 = bp (X1)Γ1 bq (X2)Γ2 . p+q=n

Property 6. If X is connected, then:

( 1 (2) |Γ| |Γ| ﬁnite; and b (X)Γ = 0 0 otherwise.

Look at n C1(X) = ZΓ → C0(X) = ZΓ → H0(X) → 0 1 0 and use homotopy invariance. 1-cells have ei 7→ e (gi −1). Then for the ﬁnite group case, H0(X) = Z, which as a ZΓ-module has dimension 1/|Γ|. In the inﬁnite case we get

n d N Γ −→N Γ → H0(X) → 0 and we claim that the dimension of the image of d is 1.

(2) Remark 5.20. Also a limiting process for residually ﬁnite groups can express bi as weighted limit of the usual Betti numbers.

Example 5.21. These properties allow us to deduce the L2-Betti numbers of some spaces.

1 (2) (2) • S with group Z: b0 = b1 = 0 from connectedness, inﬁnite group and χ formula (or duality).

• This does the torus too by Property 5.

(2) (2) (2) • For a surface Σg and universal cover, b0 = b2 = 0, b1 = 2g − 2. W 1 • For the free group Fn (so the space n S ; we talk of the Betti numbers of a group, by which we mean the Betti numbers of a corresponding K(π, 1)), we have b0 = 0, b1 = n − 1 • For a knot complement, we need to ﬁnd out, but guess that they are all zero. L2-torsion should be the Alexander polynomial. Also, for a 3-manifold we guess all Betti numbers are zero. L2-torsion is hyperbolic volume or Gromov norm.

43 5.6 Application to Manifolds 4k Definition 5.22. Let X be a compact oriented manifold, and ρ: π1(X) → Γ a representation. Then define (2) (2) (2) σΓ (X, ρ) = σΓ (λX ) ∈ R (2) where λX is a Hermitian form defined as the following intersection form: (2) ∼ 2k λX : H2k(X; N Γ) → H2k(X, ∂X; N Γ) = H (X, N Γ) → HomN Γ(H2k(X; N Γ), N Γ). The last map may not be an isomorphism: instead of Universal Coefficient Theorem there is a spectral sequence. The definition of the map is the usual one. Remark 5.23. To define σ for hermitian form h on a projective module P which is not a free n module, use the inverse Q in K0, so that P ⊕ Q = (N Γ) and take σ(hP ⊕ 0Q) as the definition of the signature. Definition 5.24 (Lück). Given M a finitely generated module over N Γ, define the torsion of M

T (M) = {m ∈ M : f(m) = 0 for all linear f : M → N Γ}.

(2) The usual definition of the torsion submodule fails for non-commutative rings. Now, λX is viewed as a pairing on H2k(X; N Γ)/T (H2k(X; N Γ)). Theorem 5.25 (Lück). M/Torsion is projective. (Like f.g. abelian groups.) That is, any M is of the form M/T (M) ⊕ T (M). Theorem 5.26 (L2-signature theorem (Atiyah)). If Y 4k is a closed manifold then for any representation ρ: (2) σΓ (Y, ρ) = σ(Y ) Actually, this is true for any elliptic operator. e.g. lift a twisted Dirac operator to the Γ-cover, compute the index via dimN Γ. The above theorem is also true for Poincarécomplexes. 4k ρi ∼ Lemma 5.27. If π1(Xi ) −→ Γ are given (additively) such that ∂X1 =φ ∂X2 and

' π1(∂X1) / π1(∂X2) ρ2 qqq ρ1 qq qqq qqq Γ xq then (2) (2) (2) σΓ (X1, ρ1) + σΓ (X2, ρ2) = σΓ (X1 ∪ X2, ρ1 ∪ ρ2). Proof. The proof is exactly the same as the usual proof of Novikov Signature additivity: decompose the module as a projective and torsion part, ignore the torsion and follow through the deﬁnitions.

44 2 (2) (2) Corollary 5.28. The reduced L -signature σeΓ (X, ρ) = σΓ −σ(X) is an invariant of the boundary ∂X and the representation ρ|∂X . Remark 5.29. Given (M 4k−1, g, ρ), where g is a metric we can deﬁne the η invariant, which is independent of the choice of metric:

(2) (2) ηeΓ (M, g, ρ) = ηΓ (M, g, ρ) − η(M, g) (Theorem of Gromov and Cheeger). By Atiyah-Patodi-Singer,

(2) (2) ηeΓ (M, ρ) = σeΓ (X, ρ) if M = ∂X. Thus, the reduced L2 signature is deﬁned when there does not exist a null-bordism. The η-invariant is obtained using the eigenvalues of the σ-operator.

3 (2) Deﬁnition 5.30. Given a knot K ⊆ S , and a representation ρ: π1(MK ) → Γ, deﬁne σeΓ (K, ρ) = (2) ηeΓ (MK , ρ). ∼ Example 5.31. Γ = Z, ρ = abelianisation π1 → H1 = Z. Then we may as well choose an expedient 4-manifold with which to make our calculations: (2) (2) σeΓ (K, ρ) = σΓ (W0, ρ) ∼ where W0 is the nice 4-manifold we constructed earlier with π1(W ) = Z which is bounded by MK . This is just σ(2)(λ(2)) = σ(2)(λ(2)) which by Corollary 5.17 is eΓ W eZ W Z σZ(K) dz. S1 6 Casson-Gordon Invariants and the Cochran-Orr-Teichner Fil- tration

2 (2) To summarise the previous section, we deﬁned the reduced L -signature σΓ (K, ρ) ∈ R, given 3 ρ: π1(SK ) → Γ, such that, if Γ = Z and ρ is the abelianisation, we have that Z (2) σe (K, ρ) = σz(K) dz. Z S1 What we really want are more interesting choices of Γ, ρ, which are such that these are concordance invariants. We already know that for Γ = Z we can detect the non-algebraically slice knots. Recall that we are aiming to reproduce the famous result of Casson-Gordon that the twist knots are not slice, using the L2 signatures to detect the intersection forms on the 4 manifold. We will choose a 4-manifold with a more complicated fundamental group, and correspondingly more complicated representations; these will be related to quotients π/π(n) of π by its derived subgroups.

45 Definition 6.1. Define Q(t) Γ = o Z Q[t±1] ±1 where Z = hti; define A := Q(t)/Q[t ], so Z acts on A via right multiplication by t. Γ is a metabelian group. A is an infinite dimensional Q-vector space. The construction of A is analogous to the situation where Q is the field of fractions of Z, and the Q/Z is formed to be the value ring −1 of linking forms. Q/Z was for torsion abelian groups, whereas A is for torsion Q[t, t ] modules. There are no finite dimensional representations of Γ, except for 1-dimensional ones which factor 2 through Z, so we are going to have to use l Γ to make our interesting representations. Now, define a representation variety

Rep∗(π, Γ) := {ρ: π → Γ | ρ(meridian) = t}.

Note that since Γ is metabelian they will all factor through π/π(2), where the (2) here denotes the 2nd derived subgroup. (In fact, Γ(1) = [Γ, Γ] = A.) Further, we know that:

π π(1) π = o , π(2) π(2) π(1) since the sequence (1) π π π ∼ 0 → → → = Z → 0 π(2) π(2) π(1) (1) is exact and splits since is free. Therefore, Rep∗ is given by ρ id: π → A , where ρ Z o π(2) o Z o Z (1) must be a homomorphism of abelian groups π → A commuting with the -action. π(2) Z Deﬁnition 6.2. π(1) AK = ⊗ Q π(2) the rational Alexander module of K.

∗ Thus, Rep (π, Γ) = Hom −1 (A ,A) is a finite dimensional -vector space. (We could work Q[t,t ] K Q with integers here, but it is easier not to.) The dimension is finite because AK is a finitely generated −1 torsion module over Q[t, t ], as we will prove shortly. It should be reasonably clear, however, from the fact that the Alexander polynomial is always monic over Q, and this represents the zero element in AK . Therefore any high powers of t can always be replaced with lower ones using the relation imposed on AK by the Alexander polynomial. Theorem 6.3 (Cochran-Orr-Teichner ). If K is slice then there is a vector subspace R ⊆ Rep∗(π, Γ) (2) of half dimension such that σΓ (K, ρ) = 0 for all ρ ∈ R. In fact, R comes from a Lagrangian of the Blanchfield form.

46 Proof. We denote by M the 0-framed surgery along K ⊆ S3, and by M its infinite cyclic cover. (1) We also abbreviate π = π (M). Then π (M) = π(1), and H (M) = π . By definition A = 1 1 1 π(2) K (1) π ⊗ ∼ H (M; ). To prove that it is a f.g. torsion module over [t, t−1], we use the Gysin π(2) Q = 1 Q Q sequence, with Q coefficients. (As always from now on, we would like to not have to use Q but it makes life a lot easier.) The sequence which arises, where we use rational coefficients, is:

1−t b 1−t H1(M) −−→ H1(M) → H1(M) −→ H0(M) −−→ H0(M) ∼ ∼ M is connected so H0(M) = Q, and the map (1 − t) here is the zero map. Also H1(M) = Q, and the map labelled b is an isomorphism, since any surjective homomorphism from Q to Q is an isomorphism. Therefore, by exactness of the sequence, the map (1 − t): H1(M) → H1(M) is a −1 surjection. This cannot be the case if there is a free part of H1(M) over Q[t, t ], since 1 − t is not invertible. Therefore it is indeed a torsion module. It is finitely generated because 1-chains are −1 f.g. free, Q[t, t ] is a Principal Ideal Domain, and so is Noetherian, so a sub-quotient is finitely generated. ∼ Note that the same argument would apply to show that for any CW complex with H1(·; Q) = Q, −1 the 1st homology of the infinite cyclic cover with Q coefficients is a f.g. torsion Q[t, t ] module. We now recall the algebraic approach to the torsion linking form on the homology AK = −1 H1(M; Q) = H1(M; Q[t, t ]) of the infinite cyclic cover: the Blanchfield form. ∼ 2 −1 First, apply Poincaréduality with twisted coefficients to get AK = H (M; Q[t, t ]). From the coefficient sequence −1 Q(t) 0 → Q[t, t ] → Q(t) → → 0 Q[t, t−1] we get the Bockstein homomorphism, which fits into the long exact sequence which arises from this short exact sequence:

1 1 Q(t) 2 −1 2 H (M; Q(t)) → H (M; ) → H (M; Q[t, t ]) → H (M; Q(t)). Q[t, t−1]

−1 −1 Now, Q[t, t ] is a principal ideal domain, so its quotient field Q(t)) is flat over Q[t, t ]. Therefore, ∼ −1 H1(M; Q(t)) = H1(M; Q[t, t ]) ⊗ Q(t) by the universal coefficient theorem for homology. Since ∼ 1 Q(t) is a field, H1(M; Q(t)) = H (M; Q(t)). With Poincaréduality, and the fact that AK is −1 Q[t, t ]-torsion, we see that both of the end groups displayed in the previous sequence are zero, ∼ 1 Q(t) meaning that AK = H (M; −1 ). Another application of a universal coefficient theorem yields: Q[t,t ]

∼ ∧ Q(t) AK = AK := Hom [t,t−1](H1(M; Q); ) Q Q[t, t−1] which isomorphism is the adjoint of the Blanchﬁeld form:

Bl : AK × AK → A

47 ∧ a unimodular hermitian linking form. (AK is sometimes called the Pontrjagin dual.) Geometrically, do the usual linking form construction, multiplying an element of H1(M; Q) by polynomials until it is zero, choosing a surface whose boundary is this chain, intersecting this with the other element, and then dividing out by whatever we needed to make the 1st 1-cycle bound. Geometrically, it is then clear that this is a Hermitian form. The theorem constructs a Lagrangian

∗ ∧ R ≤ Rep (π, Γ) = Hom −1 (A ,A) = A ∼ A Q[t,t ] K K = K where the last isomorphism is by the Blanchﬁeld isomorphism as above. We now show that there exists a Lagrangian R. Let W be B4 minus a slice disk for K, so that 3 3 ∼ ∂W = SK = M, the zero surgery on S along K, and H1(W ) = H1(M) and H2(W ) = 0. (So homologically slice will be enough in this case to prove the theorem, i.e. a knot which is slice in a homology 4-ball.) Consider the following sequence of homology of covering spaces:

0 =∼ → H2(W ) ,→ H2(W, M) → H1(M) → H1(W ) H1(W, M) −→ H0(M) −→ H0(W )

Deﬁne R := im(H2(W, M)) → H1(M) = ker(H1(M) → H1(W )) Then we claim:

(1.) 1 dim (R) = dim H (M) Q 2 Q 1 (2.) If ρ: π → Γ comes from R then it extends to ρe: π1(W ) → Γ. The point here is that we can calculate the L2 signature of (M, ρ) via the 4-manifold with its (2) (2) extended representation ρ: σeΓ (K, ρ) = σΓ (W, ρe) − σ1(W ). The ﬁnal step will then be to show (2) (2) that the Betti numbers vanish: b2 (W, ρe) = 0 ⇒ σΓ (W, ρe) = 0. This is Lemma 6.5 below. To see (1.), we need to look at linking forms on W . They are constructed just like the linking form above:

±1 ∼ 4−i ±1 ∼ 3−i Q(t) Hi(W ) = Hi(W ; Q[t ]) = H (W, M; Q[t ]) = H (W, M; ) Q[t±1]

∼ ±1 Q(t) = Hom [t±1](H3−i(W, M; Q[t ]), ) Q Q[t±1] Again this uses Poincar´eDuality, the Bockstein homomorphism, which is again an isomorphism since all are torsion modules again. The universal coeﬃcient isomorphism follows again since there ∼ ∧ are no higher Ext groups; we therefore have isomorphisms Hi(W ) = H3−i(W, M) . This yields the following diagram:

∂∗ i∗ 0 / H2(W ) / H2(W, M) / H1(M) / H1(W ) / H1(W, M) / 0

=∼ =∼ =∼ =∼ =∼ ∧ ∧ ∧ ∧ ∧ 0 / H1(W, M) / H1(W ) / H1(M) / H2(W, M) / H2(W ) / 0 i∧ ∂∧

48 ∧ Noting that dimQ N = dimQ N, we see that the dimensions on the top and bottom sequences agree, so the standard argument for Lagrangian dimensions applies. To see the extension of the representation, x ∈ R maps to a function Bl(x, •), which comes from ∧ an element of y ∈ H1(W ) which we can use to extend ρ:

π1(W ) (2) = H1(W ) o H (M) π1(W ) o Z 1 o Z ρ lll y id lll o lll lll ulll Γ = A o Z R is a Lagrangian of the Blanchfield form. To see this, take x, x0 ∈ R. Now, we can lift x to an ∧ element w ∈ H2(W, M) such that linking with x, which is an element Bl(x, •) ∈ H1(M) coincides ∧ 0 with linking with w, which is an element y ∈ H1(W ) . We can think of x ∈ H1(M) as an element ∧ of H1(W ) under the inclusion induced map i∗. We have Bl(x, •) = Bl(∂∗(w), •) = i (BlW (w, •) by ∧ commutativity of the diagram above. Then by definition of i , we get that Bl(x, •) = BlW (w, •)◦i∗, 0 0 0 0 so that Bl(x, x ) = BlW (w, i∗(x )). However, x also belongs to R, so i∗(x ) = 0 ∈ H1(W ) which implies Bl(x, x0) = 0 by linearity. The proof is then finished modulo Lemma 6.5.

Remark 6.4. The basic idea is that the inﬁnite cyclic cover of M behaves like a surface, and the slice disk complement W behaves like a null-cobordism of the surface. This is true for ﬁbred knots, while for other knots it is true homologically.

Lemma 6.5. Let Γ be a Poly-Torsion-Free-Abelian (PTFA) group. This means a group which can be constructed with a ﬁnite number of iterated extensions by torsion-free abelian groups. If 4 ∼ ∼ (2) H1(W ; Q) = H1(∂W ; Q) = Q and ρ: π1(W ) → Γ then bΓ = b2(W ). 3 (n) Remark 6.6. If π = π1(S \ K) is a knot group, then π/π is PTFA; Strebel showed that the quotients are torsion free. Also, if Γ is torsion free, then ZΓ has an Ore skew-ﬁeld of fractions. −1 −1 KΓ: it has no zero divisors and all pairs as can be reversed to pairs of the form c d. ZΓ injects into KΓ. An example of a sequence of PTFA groups is as follows. These groups are universal with respect to maps of knot groups into solvable groups.

Q(t) KΓ0 KΓ1 Γ0 = Z ← Γ1 = ±1 o Z = o Γ0 ← Γ2 = o Γ1? Q[t ] Q[Γ0] Q[Γ1] Note that A is torsion-free abelian. The final group is a first attempt at an extension by a value ring for a secondary Blanchfield pairing. It is not quite right yet however. Note that we used above the fact that Q[Γ0] is a PID. This is not true for Q[Γ1], so we make a small alteration: we include Q[Γ1], which is actually a twisted polynomial ring, into a larger ring where the coefficients of the original polynomial ring are inverted. KΓ0 ±1 KΓ0 ±1 QΓ1 = Q [t ] ⊆ K Q [t ] Q[Γ0] α Q[Γ0] α

49 We have here twisted polynomial rings: α is a choice of automorphism of the ring of which it is a subscript, which occurs when commuting coeﬃcients and the indeterminate t. Due to this choice, the identiﬁcations here with twisted polynomial rings are non-canonical. We now have a Principal Ideal Domain. Γ2 is therefore given by:

KΓ1 o Γ1 (1) ±1 K[Γ1 ]α[t ]

(1) Note that the denominator here is the same as above i.e. Γ1 = [Γ1, Γ1], the commutator subgroup KΓ0 of Γ1, is isomorphic to [ ]. Q Q[Γ0] We don’t really want to have to do this, but having a Principal Ideal Domain at certain points turns out to be very useful, particularly when defining non-singular higher order Blanchfield pairings, but also in order to sidestep some slice-ribbon difficulties to do with R ⊆ R⊥ versus R = R⊥, by making the functor •∧ right exact. Iterating the process, we define:

KΓn Γn+1 = o Γn (1) ±1 K[Γn ][t ] the “universal rationally solvable groups” based on Z = Γ0, corresponding to H1 of a knot comple- n ment (we could try taking Γ0 = Z to deal with link complements). Theorem 6.7 (Main Theorem of Cochran-Orr-Teichner). If K is slice then it is h-solvable for all 1 h ∈ 2 N. Deﬁnition 6.8. A knot is (0)-solvable if and only if its Arf invariant is zero. This implies the existence of a well-deﬁned obstruction to being slice:

0 0 B0 ∈ L (KΓ0)/L (ZΓ0) which can be detected using L2 signatures (different use of the letter L) and therefore corresponds 1 to integrating the signature σz(K) over the circle z ∈ S . This is the usual obstruction to a knot being algebraically slice: the Blanchfield or Seifert form. This gives the normal twisted signature: −1 T B0 is essentially of the form (1 − z)S + (1 − z )S , a hermitian matrix over the group ring ZΓ0, which becomes invertible upon localisation, that is, when we consider it as a matrix over the quotient field KΓ0. This then gives us an element of L-theory: a symmetric non-degenerate form. 0 Quotienting by L (ZΓ0) corresponds to a lack of signature at z = 0. A knot is (0.5)-solvable if and only if K is algebraically slice ⇒ B0 = 0. (the actual definition of h-solvable is in terms of gropes or in terms of intersection forms on the middle dimensional homology of derived covers of a 4-manifold whose boundary is M. A knot is (1)-solvable implies there exists a Lagrangian P0 inside B0 such that for all f0 ∈ P0 there is a well defined obstruction

0 (2) L (KΓ1) σeΓ B1 ∈ 0 −−→ R L (ZΓ1)

50 That is, there is a representation extending to a 4-manifold which defines an intersection form. This doesn’t depend on the choice of extensions. If a knot is (1.5)-solvable, then the Casson-Gordon invariants vanish. (1.5) solvable implies that there exists P0 ∈ B0 such that B1(p0) = 0 for all p0 ∈ P0. This implies there is a Lagrangian P1 ∈ B1(p0). In general, a knot being (n)-solvable implies that there exists P0 such that for all p0 ∈ P0, there exists P1(p0) ∈ B1(p0) such that for all p1 ∈ P1, there exists P2(p0, p1) ∈ B2(p0, p1) such that for σ(2) 0 0 eΓn all p2 ∈ P2, ... there exists a well-defined obstruction Bn(p0, .., pn−1) in L (KΓn)/L (ZΓn) −−→ R. (n.5)-solvable means that the knot is (n)-solvable and Bn = 0. Theorem 6.9 (Cochran-Orr-Teichner, Cochran-Harvey-Leidy). There exist knots which are (n)- solvable but not (n.5)-solvable for all n ∈ N. Remark 6.10. Before embarking on the proof of Lemma 6.5, we make some remarks. We will actually prove that dimKΓ(H2(W ; KΓ)) = b2(W )) in a homological algebra sense rather than in the L2 way. Why is this? Recall that Γ is PTFA: we can do a completion QΓ ⊆ N Γ. Any Von Neumann algebra satisfies the Ore condition, with no zero-divisors, so has an embedding N Γ ⊆ UΓ. This last is not a skew field, but any finitely presented module over it is projective, and quite nice. For example, if Γ = Z ∞ 1 then N Γ = L (S ; C) is all bounded functions and UΓ is all measurable functions (e.g. 1/x for x 6= 0 is now allowed). We can then do a division closure of QΓ ⊆ KΓ by closing in UΓ. This closure actually is the Ore localisation if that exists. Now, the Ore localisation is flat, so that:

dimΓ(H2(W ; N Γ)) = dimΓ(H2(W ; N Γ) ⊗N Γ UΓ).

We can extend the dimension function to UΓ but not the trace function. However, dim = trΓ : K0(N Γ) → R does extend to K0(UΓ) → R (K0 denotes the projective class group of sta- n ble isomorphism classes of modules over a certain ring) because any projection p:(UΓ) Q satisfies p2 = p · p∗ which makes it bounded because of normal algebra. By flatness, dimΓ(H2(W ; N Γ) ⊗N Γ UΓ) = dimΓ(H2(W ; UΓ)). Also, H2(W ; KΓ) ⊗KΓ UΓ = H2(W ; UΓ). The dimension is an integer over a skew-field. (This shows the Atiyah conjecture for PTFA groups: KΓ is the Ore localisation, and this shows that L2 Betti numbers are integers.) Thus H2(W ; KΓ) is free (over a skew field), which implies that so is H2(W ; UΓ), and so dimΓ(H2(W ; UΓ)) ∈ Z. Γ Proof of Lemma 6.5. We aim to prove that b2 (W ) := dimKΓ(H2(W ; KΓ)) = b2(W ). We claim that Γ Γ Γ Γ b0 = b1 = b3 = 0 for W . Then this implies the lemma since the Euler characteristics agree χ = χ. Firstly, note that with a skew-field the universal coefficient spectral sequence collapses and we have a universal coefficients isomorphism:

i ∼ H (X; KΓ) = HomKΓ(Hi(X; KΓ), KΓ)

Γ Γ Poincar´eduality therefore means that it is enough to show that b0 = b1 = 0.

51 First, KΓ H (W ; KΓ) = 0 {(1 − g)k | g ∈ Γ, k ∈ KΓ} the set of co-invariants for Γ-actions on KΓ, as usual for a connected complex. For g 6= 1, 1 − g is Γ invertible so this is zero. Now, in order to show the remaining fact, which is that b1 = 0, we need another Lemma.

Lemma 6.11. Let X be a connected finite CW complex, ρ a representation π1(X) → Γ, which is Q KΓ non-trivial, to a PTFA group. If b1 (X) = 1 then b1 (X) = 0. (Recall that for a PTFA group KΓ exists and ZΓ ⊆ KΓ.) 1 1 Proof. Let f : S → X be a map which induces an isomorphism on H1(X; Q). Then Hi(X,S ; Q) = Γ 0 for i = 0, 1. Let Xb −→ X be the covering space with deck group Γ, and form the pull back covering: 1 1 1 Sc → S . Then look at the chain complex C∗(X,b Sc; Q). It is a free QΓ chain complex such that 1 Hi(C∗ ⊗QΓ Q) = 0 for i = 0, 1, because the latter is just Hi(X,S ; Q). We claim that this implies that Hi(C∗ ⊗QΓ KΓ) = 0 for i = 0, 1. From this claim, the lemma follows because then the relative homology groups being zero imply that the absolute groups 1 1 1 Hi(X; KΓ) = Hi(Xb; Q) = Hi(Sc; Q) = Hi(S ; KΓ), and the chain complex of S is given by z−1 1 1 f∗ ρ QΓ −−→ QΓ where z is the image of 1 ∈ π1(S ) under the map π1(S ) −→ π1(X) −→ Γ. Now, 1 z − 1 is invertible in KΓ and so the complex is contractible and so has Hi(S ; KΓ) = 0 for i = 0, 1, proving the lemma. Now, to see the claim we shall use the following theorem of Strebel: 0 Theorem 6.12. If Γ is a PTFA group and f : F → F is a homomorphism of free QΓ-modules such that f ⊗ Q is injective, then f is injective. −1 f −1 Example 6.13. Γ = Z, Q[t, t ] −→ Q[t, t ]; if f(1) 6= 0 in Q i.e. f ⊗ Q 6= 0, then multiplication −1 n f −1 by f is injective. But Q[t, t ] −→ Q[t, t ]; just need to show that det 6= 0 but this is the case because det(f(1)) 6= 0. In the general case with a non-commutative Γ, we must iterate this kind of argument. The consequence of this theorem for our purposes is that, if the rank of F and F 0 are equal, then f ⊗ Q is an isomorphism, and therefore so is f ⊗ KΓ. This is because KΓ is flat over QΓ, so the tensor product preserves injectivity of maps. With Q or KΓ coefficients we then have vector spaces of the same dimension, so injective maps are isomorphisms. (This is the same as saying that QΓ embeds in its Cohn localisation, which is the universal localisation with respect to the diagram

QΓ / KΓ tt9 ttt ttt tt Cohn(QΓ) where both the maps originating from QΓ are injections.

52 Continuing with the proof of the claim, let C∗ be a free chain complex over QΓ such that Hi(C∗ ⊗QΓ Q) = 0 for i = 0, .., n (n = 1 in our case). Since each of the modules in C∗ ⊗ Q are f.g. projective (in fact free), there are partial chain homotopies h: Ci ⊗ Q → Ci+1 ⊗ Q which are chain contractions for C∗ ⊗ Q up to dimension n, that is as far as the complex is acyclic. Since C∗ is free, these partial homotopies can be lifted. Furthermore ∂h + h∂ is the identity up to dimension n, and so the lifted homotopies are also isomorphisms, by the consequence of Strebel’s result above. This means that C∗ ⊗ KΓ is also chain contractible up to dimension n, which proves the claim as required and ﬁnishes the proof.

Example 6.14. Now let’s give some example calculations and an application. Recall the twist knots Kn, with Seifert matrices: −1 1 0 n Recall that they are algebraically slice if and only if 4n + 1 is a square which is if and only if n = m(m + 1) ≥ 0. Also K0,K2 are actually slice. Let s, l be the curves on the Seifert surface which go round each of the generators of H1(F ). The curves γ1 = −m.s + l, γ2 = (m + 1).s + l are + the only primitive curves on F with lk(γi , γi) = 0. See Kauﬀman p.223 “On Knots” where these curves are seen to be torus knots T (m, m + 1). Thus m = 0, 1 cases are unknots and so K0,K2 are slice. The other cases are not unknots: we use this to show that the knots are not actually slice (though they are algebraically slice).

Theorem 6.15. The knots Kn, n = m(m + 1), m ≥ 2 are Z-independent in the group C1. Theorem 6.16 (Cochran-Orr-Teichner, but see also Casson-Gordon and Gilmer). Let K be an algebraically slice genus one knot with non-trivial Alexander polynomial. If K is slice then (2) + σ (γi) = 0 for one of the two curves γi on a Seifert surface which have lk(γ , γi) = 0 Z i Remark 6.17. The second theorem implies that the knots of the ﬁrst are not slice by checking that the signatures σ(2)(T (m, m + 1)) 6= 0 for the torus knots T (m, m + 1). There is a bit more Z work to show independence.

Proof of COT/CG. Let F be a genus 1 Seifert surface for K. The curves γi in H1(F ) are La- grangians of the Seifert form. They don’t generate H1(F ) over Z though. Recall that:

−1 λ ∗ −1 H1(F ) ⊗ Z[t, t ] −→ H1(F ) ⊗ Z[t, t ] A → 0

−1 −1 T where A = H1(MK ; Z[t, t ]) is the Alexander module of K, and λ = (1 − t)S + (1 − t )S is a hermitian restriction of the Blanchfield form. (i.e. Bl(a, b) = λ−1(a)(b) where λ is invertible over −1 Q(t) and we take the result modulo Q[t, t ].) So a Lagrangian in S gives a Lagrangian of the Blanchfield form; and vice versa actually: (γ1, γ2) ↔ (l1, l2) ⊆ L1 × L2. Define, for each i = 1, 2:

π Q(t) ρ(li): π1(MK ) = π = AK o Z → o Z = Γ1 π(2) Q[t, t−1]

53 (x, n) 7→ (Bl(li, x), n)

Remember that AK ⊗Z Q = Q ⊕ Q, and 0 6= li ∈ Li ⊗ Q = Q We have a lemma: Lemma 6.18. (2) (2) σ (K, ρ(li)) = σ (γi) eΓ1 Z These signatures are independent of the choice of li ∈ Li. In this setting, the main theorem, here the fact that K is (1.5)-solvable, now says that K being slice implies that there exists a Lagrangian L ⊆ Bl such that for all l ∈ L, σ(2)(K, ρ(l)) = 0. Here 0 eΓ1 (2) there are just two Lagrangians, so K slice implies that σ (γi) = 0 for one of the γi. This can be Z calculated not to be the case for the torus knots T (m, m + 1), obstructing sliceness of the twist knots as claimed. The proof just requires the proof of Lemma 6.18.

Proof of Lemma 6.18. Recall the picture of W0, with boundary MK in Figure 7. We define a new 4-manifold W with the same boundary: Let K be our knot and let k be the (primitive, so it can be embedded) curve on the Seifert surface F which has self linking number zero. Since it is primitive, which means that it is not a multiple in H1(F ; Z), we can think of it as being one of the bands in the Seifert surface - it may well be necessary to make an isotopy of the knot and the Seifert surface in order for this to be clear. We call W0 from Theorem 4.11 with respect to a knot J the manifold W (J). We define W by drawing W (K), and then letting the curve for k, one of the “a” curves from our Kirby diagram Figure 7 for W0, be the 1-handle for the drawing of W (k). See Figure 10. The construction is reminiscent of gropes: we are trying to test whether the secondary knot k is slice. To show that this has the correct boundary, first go through the Kirby moves for the secondary Seifert surface handles for k, to give the original k to attach a 2-handle to. Now, let l be the linking of k and a dual band γ, and let B be the twisting of γ. Then the Seifert matrix of K is:

0 l + 1 S = l B

The Alexander polynomial of K is then ±((l + 1) − tl)(l − t(l + 1)). We now compute π1(W ). It has generators which we call M, m, the meridians of K, k respectively. (hMi will be the Z in the ±1 representation to Γ1 = Q(t)/Q[t ] o Z now.)

l π1(W ) = hM, m | m[M, m ]i

This is computed because of the curve in the Kirby diagram of W corresponding to the γ band of the Seifert surface with self linking B; the “b” curve in Figure 7 and Figure 10. Choosing a base-point and following it around we can see that it links the 1-handles in the word claimed as the relation of π1(W ); the ﬁrst m corresponds to the linking which we always have in order to get the right intersection form. The ml terms appear from linking which is occurring in the box; we know the linking number from the Seifert matrix and this is the only information we need.

54 “b” curve, corresponding to dual band.

The diagram of W(k) (may have higher genus)

“a” curve, the 0 self B linking curve on F becomes 1-handle giving generator m.

The diagram of the original W(K)

1-handle giving generator M

Unspecified Twisting and linking occurs in the large box - but we know from the Seifert matrix that the “b” curve links the m 1-handle with linking number l.

Figure 10: Drawing another four manifold W with more complicated fundamental group; more sophisticated 4-manifolds are required to obstruct sliceness at the higher levels of the Cochran-Orr- Teichner ﬁltration.

−1 It links twice with the M 1-handle in opposite directions. Hence m ∈ [π1(W ), π1(W )] and so m ∈ [π1(W ), π1(W )]. There’s a short exact sequence of groups

(1) (1) ∼ π1(W ) π1(W ) π1(W )/π1(W ) = Z (1) (1) ∼ where the fact that m ∈ π1(W ) means that π1(W )/π1(W ) = Z as generated by M. Since Z is free, this sequence splits and we see that

∼ (1) ∼ π1(W ) = π1(W ) o Z = (Normal closure of m) o hMi. Now, we work modulo the second commutator group, so all conjugates of m commute. We claim

55 that: π (W ) [t, t−1] 1 ∼ Z hMi (2) = o π1(W ) lt − (l + 1) To see this, modifying the above short exact sequence yields

(1) (2) (2) (1) ∼ π1(W ) /π1(W ) π1(W )/π1(W ) π1(W )/π1(W ) = Z which means that ∼ π1(W ) = H1(W ; Z[Z]) o hMi −1 where H1(W ; Z[Z]) = H1(W ; Z[t, t ]) is the homology of the universal abelian cover of W , with M acting on it as deck transformations: conjugation by M translates as multiplication (on the right) −1 by t. This homology is generated by m; this becomes the generator of Z[t, t ], as M generates the deck group of the Z-covering of W , there is one copy of the 1-handle m for each power of t. Again the b curve yields the relation: it links m l times, passes round M, and then links m a total of l + 1 times, here with coeﬃcient t since it passed round M, before passing back round M to the original sheet of the covering, completing the loop, and completing the derivation of the claim for (2) π1(W )/π1(W ) . So there’s a commutative diagram:

π (M ) Γ 1 K 8/ 1 qqq O ρ qqq qqq qqq qq ±1 Z[t ] π1(W ) / {lt−(l+1)} o Z

(2) Note that m ∈ π1(W )/π1(W ) maps to a non-zero element of Γ1. Therefore:

σ(2)(K, ρ) = σ(2)(W 4, ρ) eΓ1 eΓ1 0 We will now calculate the right hand side via computation of H2(W ; N Γ). Let Γ := im(ρ) ≤ Γ1. We have hmi = Z ≤ Γ0. We need the following crucial Proposition to complete the argument.

Proposition 6.19. Consider a subgroup Γ0 ≤ Γ1. Then there are commutative diagrams

- 0 - 0 N Γ0 N Γ1 L (N Γ0) L (N Γ1) Q Q Q tr Qs + tr (2)QQs + (2) Γ0 Γ1 σ σ CRΓ0 Γ1 Remark 6.20. The above proposition is fundamental to our calculations. We constructed our 4-manifold in such a way that the matrix of the intersection form over ZΓ1 contains as its entries

56 0 only linear combinations of powers of a single non-trivial group element m ∈ Γ ≤ Γ1. Since our groups Γ are torsion-free, this gives an inclusion of groups ∼ Z = hmi ≤ Γ1 to which Proposition 6.19 can be applied. Thus the L2-signature for Γ can be calculated for this particular Hermitian form as an integral over the circle of certain twisted signatures.

Proof of Proposition 6.19. A homomorphism N Γ0 → N Γ1 is given by completing

2 a ⊗ Id ∈ End(l Γ0 ⊗CΓ0 CΓ1) 2 to a bounded operator on l Γ2 for any a ∈ N Γ1. Since

h(e0 ⊗ e1)(a ⊗ Id), e0 ⊗ e1i = h(e0)a, e0i · he1, e1i = h(e0)a, e0i it follows that the first diagram commutes. For details see Theorem 3.3 of Lück’s book on L2- homology. This reference also contains the statement that tensoring an N Γ0-module with N Γ1 is a flat functor. To see that the second diagram above commutes just observe that diagonalising a hermitian matrix over N Γ0 and then tensoring up the ±1-eigenspaces to N Γ1 diagonalises the induced matrix over N Γ1. Thus the commutativity of the first diagram proves the claim.

0 By Proposition 6.19, since Γ ≤ Γ1, we get a commuting diagram:

0 0 0 L (N Γ ) / L (N Γ1) MMM MM σ(2) MM eΓ1 σ(2) MMM eΓ0 MM& R so just using the using the subgroup Γ0 gives us the correct L2-signature. We now claim that:

0 0 H (W ; N Γ ) ∼ H (W (k); N (hmi) ⊗ ±1 N Γ ) 2 = 2 0 Z[m ] (2) The intersection form on the latter group has the desired signature σe for the proof of the lemma. 0 (2) To see the claim, note that Γ ≤ π1(W )/π1(W ) . Compute the second homology geometrically; 0 the 2-handles from W0(k) give a basis in the covering space with deck group Γ . The linking of the b curve with the m 1-handle means that its attaching circle is not null-homotopic and so it ∼ doesn’t give a generator of H2(W ; N Γ). We can then compute intersection numbers using hmi = Z, which then gives us the numbers with m ∈ Γ0, hence the tensoring up as a Hermitian module. Now 0 0 0 Proposition 6.19 applies again; the map L (NZ) → L (N Γ ) is exactly tensoring up. So ultimately:

(2) (2) (2) (2) (2) σ (K, ρK ) = σ (W ) = σ 0 (W ) = σ (W0(k)) = σ (k) eΓ1 eΓ1 eΓ eZ eZ ρ Note that about surjectivity of π1 −→ Γ was not essential in our choices: using the image of ρ is enough.

57 Remark 6.21. Note on Proposition 6.19: it is rather remarkable that a signature on a smaller group extends to a bigger one. This is not usually true for invariants of L-groups. It comes because the trace function extends from N Γ1 → C to N Γ2 → C. We should warn the reader that the von Neumann algebra N Γ is not functorial in Γ. This has to do with the speciﬁc choice of the Hilbert-space l2Γ. Proposition 6.19 gives the best possible functoriality which is valid for all groups. If Γ is amenable, then the equality of the reduced and maximal C∗-algebras (which are functorial!) implies that the projection Γ → {1} induces a ∗ ∗ homomorphism of C -algebras ε : C Γ → C. For example, if Γ = Z then this is given by evaluating 1 ∞ 1 a continuous function at 1 ∈ S . This clearly does not extend to N Z = L (S ).

Proofs

Proposition 6.22. Let M be a ﬁnitely generated abelian group. Then,

1 ∼ Ext (M, Z) = Hom(Tor(M), Q/Z).

Proof. We have the exact sequence

µ ν 0 → Z −→ Q −→ Q/Z → 0

1 Since Q is injective, Ext (M, Q) = 0 and we get the sequence

∗ µ ν∗ 1 0 → Hom(M, Z) −→ Hom(M, Q) −→ Hom(M, Q/Z) → Ext (M, Z) → 0

n Now, by the structure theorem for f.g. abelian groups, we can write M = Z ⊕ Zn1 ⊕ · · · ⊕ Znq . Then n n Hom(M, Z) = Hom(Z , Z) ⊕ Hom(Zn1 , Z) ⊕ · · · ⊕ Hom(Znq , Z) = Hom(Z , Z) n since Hom(A, Z) = 0 for any torsion module A. Similarly, Hom(M, Q) = Hom(Z , Q).

This gives us

1 ∼ ∗ n ∼ Ext (M, Z) = Hom(M, Q/Z)/ im ν = Hom(M, Q/Z)/ Hom(Z , Q/Z) = Hom(Tor(M), Q/Z).

n Proposition 6.23. There is an isomorphism Φ: Tor H (X) → Hom(Tor Hn−1(X), Q/Z).

n f(y) Proof. For a function f ∈ Tor H (X), we let Φ(f) be the map x 7→ s , where x ∈ Tor Hn−1(X), and sx = dy for some s 6= 0 ∈ Z and y ∈ Sn(X). Here we use the terminology that the homology groups are calculated from the chain complex

d d · · · → Sn+1(X) −→ Sn(X) −→ Sn−1(X) → ....

58 Our map is clearly a homomorphism. To prove surjectivity, take φ ∈ Hom(Tor Hn−1(X), Q/Z). We n want to ﬁnd f ∈ Tor H (X) such that φ(x) = Φ(f)(x) ∀x. So let x ∈ Tor Hn−1(X) with sx = dy for some s ∈ Z\{0} and y ∈ Sn(X). Put f = φ ◦ d. Then f(y) = φ(d(y)) = φ(sx) = sφ(x), and so f(y) φ(x) = s = Φ(f)(x) as required.

It remains to prove injectivity. Suppose Φ(f) is the zero map for some f ∈ Tor Hn(X). This f(y) f(y) means that Φ(f)(x) = s ≡ 0 for all x ∈ Tor Hn−1(X), which means that s ∈ Z, so s divides f(y). n n−1 ∗ Since f ∈ Tor H (X) we can ﬁnd m ∈ Z and g ∈ S (X) such that mf = g◦d (i.e. mf ∈ im d ). Now mf(y) = g(d(y)) = g(sx) = sg(x)

f(y) g(x) 0 so that s = m ∈ Z for all x. But if m divides g(x) for all x, then we may write g = mg for some g0 ∈ Sn−1(X). Then mf = mg0 ◦ d ⇒ f = g0 ◦ d ∈ im d∗, so f is the zero element in Tor Hn(X).